Algebraic equations are mathematical statements where two expressions are separated by an equal sign.

The variables can be on both sides of the equality sign. We use letters to denote variables.

Examples:

$x = 4 + 2$. Here, we observe that the variable is on one side of the equation.

$2x + 3 = x + 4$. Here, we can observe that the variables are on both sides of the equation.

## Writing Algebraic Equations

An algebraic equation contains variables, numbers and an equal sign that can be interpreted into meaningful statements. These variables are also called as literal coefficients.
Any word problem can be changed into a simple and clear algebraic equation, which makes solving it much easier.

Solving a word problem depends on how well one understands it and translates it into a mathematical statement.
So, the main steps that have to be followed while dealing with a word problem are:
• Understand what is given.
• Understand what is to be found
• Understand the relation between what is to be found and what is given.
Word problems get easier if you understand the keywords that are used in the problem.
Lets consider some

 Operation Symbol Keyword Addition + Sum Added to Total More than Increased by Subtraction - Minus Less Difference Decreased by Fewer than Multiplication * Multiplied by Product of Times Of Division ÷ Out of Ratio of Per Quotient of Percent Equal = Is Are Will be Gives Were Power of ^ Squared Cubed

Algebraic equations are very useful in solving word problems as they show a symbolic way of solving them.
When reading a word problem, one must be very careful in interpreting the sentences to equations.

The following steps can be followed to convert a word problem into an algebraic equation:
• Understand the problem by reading it thoroughly
• Check what is to be found and assign it a variable, say x.
• Check what is given.
• Form an equation connecting the given values and the variable, x.
Hence, we get an equation which can be solved to get the variable.

### Examples on Writing Algebraic Equations

Given below are some examples that explain how to form an algebraic equation.

Example 1:

Half of a number increased by 5 is 10

Solution:

Let the number be x.
$\frac{1}{2}$ $x + 5 = 10$

Example 2:

Jane gets 20 dollar as pocket money. How much money does she need to buy a toy that costs 92 dollars?

Solution:

Step 1: First consider what is given. Jane has 20 dollar and wants to buy a toy that costs 92 dollars. The question is how many more dollars are needed to buy the toy.

Step2: Let us call the extra amount needed as x.

Step3: So, she already she has 20. We need x more money to get 92 dollars. Hence, an equation can be formed
$20 + x = 92$

Step4: Subtract 20 from both sides
$x = 72$
So, Jane needs 72 dollars more to buy the toy.

Example 3:

The sum of 11 and a quantity is multiplied by 2 to get 8. Find the quantity.

Solution:

Let us consider the quantity as u.
Given, the sum of 11 and a quantity is multiplied by 2 to get 8.
So, the algebraic equation formed will be as follows:
$(u + 11) \times 2 = 8$

Divide both the sides by 2
$(u + 11) = 4$

Subtract 11 from both the sides
$u = 4 – 11$
u = 7 dollars

## Solving Algebraic Equations

Algebraic equations can be solved by isolating the term containing the variable first. This can be done by adding or subtracting terms from both the sides of the equal sign. The value of the variable can be found by then dividing the whole equation by the co-efficient of the variable. The result we thus get is the solution of the algebraic equation.

Let us observe how to solve algebraic equations with the help of the following examples.

Example 1:

$4x = 3x + 1$

Solution:

$4x = 3x + 1$
$4x - 3x = 3x - 3x + 1$ (By subtracting 3x from both the sides, all the variables are on the same side of the equation.)
$x = 1$

Example 2:

$5x = 3x - 16$

Solution:

$5x = 3x - 16$
$5x - 3x = 3x - 3x - 16$ (By subtracting 3x from both the sides, all the variables are on the same side of the equation.)
$2x = 16$
$x = 8$

Example 3:

$5p =$15

Solution:

$5p = 15$
$\frac{5}{5}$ $p =$ $\frac{15}{5}$
$p = 3$

Example 4:

$3a + 5a = 4$

Solution:

$3a + 5a = 4$
$8a = 4$
$\frac{8}{8}$ $a =$ $\frac{4}{8}$
$a =$ $\frac{1}{2}$

Example 5:

$7m = 2m + 18$

Solution:

$7m = 2m + 18$
$7m + 2m = 2m + 2m + 18$ (By adding 2m on both the sides, all the variables are on the same side of the equation)
$9m = 18$
$\frac{9}{9}$ $m =$ $\frac{18}{9}$
$m = 2$

## Solving Algebraic Equations with Fractions

Algebraic equations contain terms in the form of integers, decimals or fractions. When dealing with fractions, the simplest way is to remove the fractional form. This can be done by multiplying the denominator term with the numerator on both sides of the equation. The steps will vary according to the question but the idea is the same - remove the denominator. The process of changing an equation without fractions is called the clearing of fractions.

### Examples on Solving Algebraic Equations with Fractions

Given below are some examples that explain how to solve algebraic equations with fractions.

Example 1:

$3x +$ $\frac{2}{7}$ = $5$

Solution:

Here, there is a fraction $\frac{2}{7}$. The problem would be easier if we can remove the fraction. The only way we can do this is by multiplying both sides by 7.
$7 \times (3x + $$\frac{2}{7}$$)$ = $7 \times 5$
On the left hand side, 7 can be distributed.
$7 \times 3x + 7 \times $$\frac{2}{7} = 7 \times 5 21x + 2 = 35 Subtract 2 from both the sides 21x = 33 Divide both the sides by 21 x = \frac{33}{21} Example 2: (\frac{6}{8})$$ \times (x + 3) = 9$

Solution:

This is an equation that contains a fraction $\frac{6}{8}$.
This can be solved in many ways. But, lets look at the easiest way: Clearing the fractions.
$(\frac{6}{8})$$\times (x + 3) = 9 To remove the fraction from the left hand side, multiply it by 8. Now, this process will alter the equation. So, perform the same process on the right hand side also. 8 \times$$[(\frac{6}{8})$$\times (x + 3)$$]$ $= 8 \times 9$
So, $6 \times (x + 3) = 72$

Now, Distribute 6
$6x + 18 = 72$

Subtract 18 from both the sides
$6x = 72 - 18$
$= 54$

Divide by 6 on both the sides
So, $x = 9$.

The main steps for solving such equations are:
• Choose a common denominator for the equation.
• Multiply the common denominator with every term in the equation
• Remove all the denominators and hence reduce the equation to its simplest form.

Example 3:

$\frac{(x-2)}{2}$$+$$\frac{3}{2}$ = $6$

Solution:

Here, we can see that both the terms have the same denominator 2, on the left hand side. So, that can be taken as common.
$\frac{(( x - 2)+3)}{2}$ = $6$

Multiply both sides by 2
$2 \times $$\frac{(( x - 2)+3)}{2} = 2 \times 6 So, (x - 2) + 3 = 12 Subtract 3 from both sides x - 2 = 9 Add 2 on both the sides So, x = 11 Example 4: \frac{3x}{5}$$ - $$\frac{(x+3)}{10} = -7 Solution: Here, LCM of 5 and 10 is 10. Multiply by 10 on both sides 10$$(\frac{3x}{5}$$-$$\frac{(x+3)}{10})$ = $10(-7)$
$2 \times 3x - (x + 3) = -70$
$6x - x - 3 = -70$
$5x - 3 = -70$

Add 3 on both sides
$5x = -67$

Divide both sides by 5
$x$ = $\frac{-67}{5}$

Example 5:

$\frac{(x – 2)}{3}$$+$$\frac{x}{5}$ = $6$

Solution:
Here, the denominators are different on the left hand side. So, we need a common term.
Find the LCM of 3 and 5, which is 15. Multiply by 15 on either side of the equation
$15 \times $$(\frac{(x – 2)}{3}$$ + $$\frac{x}{5}) = 15 \times 6 15 \times$$(\frac{(x – 2)}{3})$$+ 15 \times$$(\frac{x}{5})$ = $90$
$5 \times (x - 2) + 3 \times x$ = $90$
$5x - 10 + 3x$ = $90$
$8x - 10$ = $90$

Add 10 on both the sides
$8x$ = $100$

Divide 8 on both the sides
$x$ = $\frac{25}{2}$

## Factoring Algebraic Equations

Factoring an algebraic equation is just the reverse of the distribution law. So, by rule
$ab + ac$ = $a(b + c)$

The main steps for factoring the equation are:
• Find the common factor in the expression.
• Factor it out.
In general, factoring is a process by which a sum is converted into a product by removing the common term or factor and putting them outside the parenthesis. So, only if there is a common factor, can any algebraic equation be factored.

### Examples on Factoring Algebraic Equations

Given below are some examples on factoring algebraic equations.

Example 1:

$24x + 9y + 3xy$ = $0$

Solution:

Here, there are three terms $24x$, $9y$ and $3xy$. Lets factor each of them.
$24x$ = $2 \times 2 \times 2 \times 3 \times x$
$9y$ = $3 \times 3 \times y$
$3xy$ = $3 \times x \times y$
The highest common factor is 3.
So,
$24x + 9y + 3xy$ = $0$
$3(12x + 3y + xy)$ = $0$

Example 2:

$2x - 4x^{2}$ = $0$

Solution:

Here, there are two terms $2x$ and $4x^{2}$. Lets factor each of them.
$2x$ = $2 \times x$
$4x^{2}$ = $2 \times 2 \times x \times x$
The highest common factor is $2$ and $x$. Therefore, $2x$ is the highest factor that divides both the terms.
So, $2x$ can be factored out as follows:
$2x - 4x^{2}$ = $0$
$2x(1 - 2x)$ = $0$
Now, it can be solved using the zero property
As $2 \neq 0$, $x = 0$ and $1 - 2x$ = $0$
$x$ = $\frac{1}{2}$
So, the solutions are $x$ = $0$, $\frac{1}{2}$

Example 3:

$-2x^{2}y - 6xy^{2}$

Solution:

Here, the factors are
$-2x^{2}y$ = $-1 \times 2 \times x \times x \times y$
$-6xy^{2}$ = $-1 \times 2 \times 3 \times x \times y \times y$
So, the common term is $-2xy$
So,
$-2x^{2}y - 6xy^{2}$ = $-2xy(x + 2y)$

Example 4:

$x( y + 1 ) + 5(y + 1)$

Solution:

Here, there are two terms $x(y + 1)$ and $5(y + 1)$
The highest common factor is $y+1$
So,
$x(y + 1) + 5(y + 1)$ = $(x + 5)(y + 1)$

Special identities
There are certain special identities that can be used for making the factoring simple
• u2 - v2 = (u + v)(u – v)
• u2 + 2uv + v2 = (u + v)(u + v)
• u2 - 2uv + v2 = (u - v)(u – v)
• u3 + v3 = (u + v)(u2 - uv +v2)
• u3 - v3 = (u - v)(u2 + uv +v2)
• u3 + 3u2v + 3uv2 + v3 = (u +v)(u + v)(u + v)
• u3 - 3u2v + 3uv2 - v3 = (u - v)(u - v)(u - v)
There are many more, but these are the main identities.

Example 4:

$9x^{2} - 64$ = $0$

Solution:

Here, $9x^{2}$ = $(3x)^{2}$
$64$ = $8^{2}$
$9x^{2} - 64$ = $0$
$(3x)^{2} - 8^{2}$ = $0$
Here, the following identity can be used
$u^{2} - v^{2}$ = $(u + v)(u – v)$
So, $(3x)^{2} - 8^{2}$ = $(3x - 8)(3x + 8)$
So, it can be simplified as $(3x - 8) (3x + 8)$ = $0$

### Quadratic Equations:

An algebraic equation in the form $ax^{2} + bx + c$ = $0$ is called a quadratic equation, where $a > 0$. A quadratic equation can be factored in two ways. For factoring, we need two numbers say, $f$ and $g$ such that $f + g$ = $b$ and $fg$ = $c$. If the equation is in the form $x^{2} + bx + c$ = $0$, that is $a$ = $1$, then the method is pretty simple.

### Examples on Factoring Quadratic Equations

Given below are some examples that explain how to factor a quadratic equation.

Example 1:

$x^{2} + 5x + 6$ = $0$

Solution:

Find two numbers so that sum is $5$ and the product is $6$. Checking by factoring we get $2$, $3$ such that, $2 + 3$ = $5$ and $2 \times 3$ = $6$
So, we can factor it as $(x + 2) (x + 3)$
So, $x^{2} + 5x + 6$ = $0$
$(x + 2) (x + 3)$ = $0$
Hence solved.

Example 2:

$9x^{2} + 6x + 1$ = $0$

Solution:

To factor this, we need two numbers so that the sum is $6$ and the product is $9$. Let us consider $3$ and $3$. So that, $3 + 3$ = $6$ and $3 \times 3$ = $9$.
So, just split the middle term as $3x + 3x$
$9x^{2} + 6x + 1$ = $0$
$9x^{2} + 3x + 3x + 1$ = $0$
$3x(3x + 1) + 1(3x + 1)$ = $0$
So, $(3x + 1)(3x + 1)$ = $0$
Hence solved by zero property.
Any quadratic equation can be solved using quadratic formulas.

$x$ = $-b \pm \sqrt{\frac{b^{2} - 4ac}{2a}}$
Example 3:

$x^{2} -7x + 10$

Solution:

Here, $a$ = $1$, $b$ = $-7$, $c$ = $10$
So, $x$ = $-(-7) \pm \sqrt{\frac{((-7)^{2} -4(1) (10))}{(2\times1)}}$

= $7 \pm \sqrt{\frac{(49 – 40)}{2}}$

= $\frac{(7 \pm \sqrt{9})}{2}$

= $\frac{(7 \pm 3)}{2}$

So, $x$ = $\frac{(7-3)}{2}$ and $x$ = $\frac{(7+3)}{2}$
$x$ = $\frac{4}{2}$ and $x$ = $\frac{10}{2}$
$x$ = $2$ and $x$ = $5$
And so, the factors are $(x - 2)$ and $(x - 5)$.

## Conditional Equation

A conditional equation is an equation which is true for the particular value of the variable it contains. The conditional equation represents a problem. We can create as many conditional identities as we want as there is no particular rule to create conditional equations, except that they must not be inconsistent equations which do not have any values of the variable for which the equation is true.

$x^{2} + 3x = 7$ is a conditional equation which asks " What are the values of x for which the square of the value increased by three times the value is seven?" The values that must be assigned to the variable so that the equation is true is called the solution to the conditional equation. If there is more than one value for which the conditional equation is true, then they are the Solution Set for the Conditional Equation.

Consider the equation $3x + 2$ = $8$. This equation is true only when $x$ = $2$. For any other value of $x$, the equation is not true. Hence, $x$ = $2$ satisfies the equation $3x + 2$ = $8$

In many cases, we can predict the existence of a solution to a conditional equation by the concrete conditions that result in the problem. For example, when there is a problem like, " the perimeter of a triangle is 32 and one of its side is 3 cm and the second side is twice the third side. What are the lengths of the sides of the triangle?" This problem clearly says it is a triangle which gives the necessary condition and the clue to solve the conditional equation results from it. In order to find out which values of the variable satisfy a given equation, we must solve it. The values that satisfy a given equation are also referred to as the Roots of the equation.

For example, 3 is a root of the equation $x + 4$ = $7$. If two equations have the same root, they are said to be Equivalent Equations. There are a few theorems that are useful to solve a given conditional equation by transforming them into an equivalent equation.

### Theorems to transform a given conditional equation into an equivalent equation

Given below are some theorems that can help us transform a conditional equation into an equivalent equation.

Theorem 1:

The Addition or Subtraction of a finite quantity to both sides of an equation results in a new equation that is Equivalent to the original equation.

If A = B is an equation and C is any finite quantity or an algebraic expression which will be finite for any finite value assigned to its variables, then A + C = B + C and A - C = B - C are equivalent equations to the equation A = B.

This theorem helps us to transfer a term from one side of the equation to the other so that we can solve it easily.

For example, $x + 4$ = $3$

To solve this equation, we should subtract $4$ from both sides of the equation.
$x + 4 - 4$ = $3 - 4$ which is equivalent to the equation $x + 4$ = $3$
Simplifying we get $x$ = $-1$
Hence, the solution to the $x + 4$ = $3$ is $-1$.

Theorem 2:

An equation which is equivalent to the given equation can be obtained by multiplying or dividing both the sides of the equation by a finite non zero quantity.

Consider the equation A = B. For any finite non zero quantity C, AC = BC and A/C = B/C are equations which are equivalent to the equation A = B. This also helps us to solve the equation.

For example, $2x$ = $8$

This equation can be solved by dividing both the sides of the equation by 2.
$\frac{2x}{2}$ = $\frac{8}{2}$
$x$ = $4$
Hence, the solution to the $2x$ = $8$ is $4$.

### Examples on Conditional Equations

Given below are some examples that explain how to solve conditional equations.

Example 1:

Solve the following conditional equation using the properties of equality

$x - 9$ = $2x - 3$

Solution:

$x - 9$ = $2x - 3$
$x - 9 + 9$ = $2x - 3 + 9$ (Adding 9 to both sides of the equation)
$x$ = $2x + 6$
$x - 2x$ = $2x + 6 -2x$ (Adding -2x to both sides of the equation)
$-x$ = $6$
$-x \times (-1)$ = $6 \times (-1)$ (Multiplying by -1 on both sides of the equation)
$x$ = $-6$
Hence, the solution to the equation $x - 9$ = $2x - 3$ is $x$ = $-6$

Example 2:

Solve $\frac{x}{5}$$+$$\frac{3}{7}$ = $\frac{5}{6}$

Solution:

$\frac{x}{5}$$+$$\frac{3}{7}$ = $\frac{5}{6}$

$\frac{x}{5}$$+$$\frac{3}{7}$$-$$\frac{3}{7}$ = $\frac{5}{6}$$-$$\frac{3}{7}$ (Subtracting $\frac{3}{7}$ from both the sides of the equation)

$\frac{x}{5}$ = $\frac{17}{42}$

Multiply both sides of the equation by 5 so that we can eliminate the denominator of the variable.
$5 \times $$\frac{x}{5} = 5 \times$$\frac{17}{42}$

$x$ = $\frac{85}{42}$

Hence, the solution to the equation $\frac{x}{5}$$+$$\frac{3}{7}$ = $\frac{5}{6}$ is $x$ = $\frac{85}{42}$

## Infinite Solutions in Algebraic Equations

An equation is said to have infinite solutions if it passes through many points. For example, let us consider y = y. It can be seen that for any real number value, this equation is satisfied. So, it has infinite solutions.

When we consider two lines, to get infinite solutions, they will intersect at many points along the line. If it is a straight line, this will only occur if both of them are on the same line. This kind of solution is called an inconsistent solution. As it represents the same line, it is called an dependent system of equations.

### Infinite solution in a pair of equation

If any of the following conditions are satisfied, then it can be said that there are many solutions.

• If the lines are parallel and overlapping.
• When graphing, all points(x, y) are simultaneously on both lines.
• While solving for a solution, using elimination or substitution method, leads to an identity.
• Slopes are equal with the same y intercepts.
• If A/D = B/E = C/F

### Examples on Infinite Solutions:

Given below are some examples on Infinite Solutions

Example 1:

Solve $3y - x = -3$ and $6y = 2x - 6$

Solution:

$3y - x = -3$
$6y = 2x - 6$

Let us solve this by graphing

$3y - x = -3$

 x 0 -3 3 y -1 1 0
$6y = 2x - 6$

 x 0 6 3 y -1 1 0

It can be seen that both the lines are parallel and overlapping. So, it has many solutions. And so, it is inconsistent but dependent.

Example 2:

Solve $x + 3y = 1$ and $2x + 12y = 2$

Solution:

$x + 3y = 1$
$2x + 12y = 2$

Using the Substitution method:
$x + 3y = 1$ -------------- (1)
$2x + 6y = 2$ ------------ (2)

Step 1: Isolate x from the first equation
$x = 1 - 2y$

Step 2: Substitute $x = 1 - 3y$ in the second equation
$2(1 - 3y) + 6y = 2$

Step 3: Solve for y
$(2)(1) - (2)(3y) + 6y = 2$
$2 - 6y + 6y = 2$
$2 = 2$

Here, we get an identity.
So, they have many solutions. Hence, the system is inconsistent and dependent.

Example 3:

Solve $x + 5y = 12$ and $4x + 20y = 60$

Solution:

$x + 5y = 12$
$4x + 20y = 60$

While comparing this with the general form of lines Ax + By = C and Dx + Ey = F, we get,
A = 1, B = 5, C = 12, D = 4, E = 20, F = 60

Now, $\frac{A}{D}$ = $\frac{1}{4}$

$\frac{B}{E}$ = $\frac{5}{20}$ = $\frac{1}{4}$

$\frac{C}{F}$ = $\frac{12}{60}$ = $\frac{1}{4}$

So, $\frac{A}{D}$ = $\frac{B}{E}$ = $\frac{C}{F}$

Hence, there are many solutions for the given set of equations. So, the system is inconsistent and dependent.

Example 4:

Solve $y = -2 + 5x$ and $15x - 3y = 6$

Solution:

$y = -2 + 5x$
$15x - 3y = 6$

First, let us write the equation in a slope intercept form y = mx + b where, m and b are the slope and y is the intercept respectively.

So, $y = -2 + 5x$
$y = 5x - 2$
Here, slope = 5
And, y intercept = -2

$15x - 3y = 6$
$y = $$\frac{15x - 6}{3} = \frac{15x}{3} - \frac{6}{3} Here, slope = \frac{15}{3} = 5 And, y intercept = \frac{-6}{3} = -2 So, the slopes are equal with the same y intercepts. Therefore there are many solutions. Hence, the system is inconsistent and dependent. ## No Solution in Algebraic Equations An equation can either be true or false according to the value we assign to the variable. When none of the real numbers makes the equation true, we say that there exists no solution for it. Let us consider a linear equation x - 5 = 0. When we plug in x = 5, it can be seen that the equation is satisfied. Hence, it has one solution x = 5. Now, consider x + 4 = x + 6. Here, whatever value we plug in for x, it can be seen that this equation will never be true. Hence, a contradiction occurs as we assume each as a solution. And so, it has no solution ### No Solution in a Pair of Equation If any of the following condition is satisfied, then it can be said that there is no solution. • If the lines are parallel with no overlapping points. • When graphing, there is no point (x, y) that is simultaneously on both lines. • While solving for a solution, using elimination or substitution method, leads to a contradiction. • Slopes are equal with different y intercepts. • If A/D = B/E ≠ C/F Such equations with no solutions are called inconsistent equations. Such systems are called as an independent system. ### Examples on equations with no solution Given below are some examples on solving equations with no solutions. Example 1: Solve 2x + 4y = 16 and x + 2y = 4 Solution: 2x + 4y = 16 x + 2y = 4 Lets graph each equation. 2x + 4y = 16  x 0 4 8 y 4 2 0 x + 2y = 4  x 0 2 4 y 2 1 0 It can be seen that both the lines are parallel without any overlap. So, it has no solution. Therefore, it is inconsistent. Example 2: Solve x + 3y = 12 and 3x + 9y = 24 Solution: x + 3y = 12 3x + 9y = 24 Using the Substitution method: x + 3y = 12 ------------- (1) 3x + 9y = 24 ----------- (2) Step 1: Isolate x from the first equation x = 12 - 3y Step 2: Substitute x = 12 - 3y in the second equation 3(12 - 3y) + 9y = 24 Step 3: Solve for y (3)(12) - (3)(3y) + 9y = 24 36 - 9y + 9y = 24 36 = 24 Here, the equation becomes a contradiction as 36 and 24 cannot be equal. So, they have no solution. And hence, the system is inconsistent. Example 3: Solve 2x + 3y = 12 and 4x + 6y = 20 Solution: 2x + 3y = 12 4x + 6y = 20 While comparing this with the general form of the lines Ax + By = C and Dx + Ey = F, we get, A = 2, B = 3, C = 12, D = 4, E = 6, F = 20 Now, \frac{A}{D} = \frac{2}{4} = \frac{1}{2} \frac{B}{E} = \frac{3}{6} = \frac{1}{2} \frac{C}{F} = \frac{12}{20} = \frac{3}{5} So, \frac{A}{D} = \frac{B}{E}$$\frac{C}{F}$

Hence, there is no solution for this set of equations. And so, the system is inconsistent.

Example 4:

Solve $x - y = 2$ and $3x - 3y = 5$

Solution:

$x - y = 2$
$3x - 3y = 5$

First, let us write the equation in the slope intercept form y = mx + b where, m and b are slope and y intercept respectively.

So, $x - y = 2$
$y = x - 2$
Here, slope = 1
And, y intercept = -2

$3x - 3y = 6$
$y =$ $\frac{3x - 6}{3}$ = $\frac{3x}{3}$ - $\frac{5}{3}$

Here, slope = $\frac{3}{3}$ = $1$

And, y intercept = $\frac{-5}{3}$

So, the slopes are equal but y intercepts are different. So, there is no solution. Hence, the system is inconsistent.

## Two Step Algebraic Equations

Two-Step Algebraic Equations are defined as solving an equation twice, that is, one step is done two times. The reason it is called so is because there are only two operations that are needed to solve the equation.

1. The opposite of addition operation is subtraction and vice versa.
2. The opposite of multiplication operation is division and vice versa.

There are two types of such equations:
• Two-step Algebraic Multiplication Equations
• Two-step Algebraic Division Equations

### Two-step Algebraic Multiplication Equations:

The processes for solving these types of two way algebraic equations are:
• Add or subtract so that the Coefficient and Variable are on either side of the equal sign, from both sides.
• Divide each side by the coefficient of the corresponding variable.
• Solve the remaining part of the equation.
Example 1:

$7y - 6 = 36$

Solution:

To solve, the variable must be on one side and all the other numbers should be on the other sides.
$7y - 6 = 36$
Add 6 on each side
$7y - 6 + 6 = 36 + 6$
$7y + 0 = 42$
$7y = 42$
Divide by 7
$\frac{7y}{7}$ = $\frac{42}{7}$
$y = 6$

Example 2:

$1 + 10u = 56$

Solution:

$1 + 10u = 56$
Subtract 1 from both the sides
$10u = 55$
Divide by 10
$u$ = $\frac{55}{10}$
= $\frac{11}{2}$

### Two-step Algebraic Division Equations:

The process for solving these types of two way algebraic equations are:
• Add or subtract so that the Coefficient and Variable are in either side of equal sign, from both the sides.
• Multiply both the sides by the coefficient of the variable.
• Solve the remaining part of the equation.

Example 3:

$\frac{y}{5}$ $- 6 = 34$

Solution:

To solve, the variable must be on one side and all other numbers to the other sides.
$\frac{y}{5}$ $- 6 = 34$
Add 6 on either side
$\frac{y}{5}$ $- 6 + 6 = 34 + 6$
$\frac{y}{5}$
$- 0 = 40$
$\frac{y}{5}$ = $40$
Multiply by 5
$\frac{5y}{5}$ = $5 \times 40$
$y = 200$

Example 4:

$\frac{-u}{6}$ $+ 1 = 13$

Solution:

$\frac{-u}{6}$ $+ 1 = 13$
Subtract 1 from both the sides
$\frac{-u}{6}$ = $12$
Multiply by -6 on both the sides
$u = 12 \times (-6) = -72$

## One Step Algebraic Equation

Any algebraic equation that requires only one set to solve can be defined as one step algebraic equation.
To solve an algebraic linear equation means to find an appropriate value for the variable so that the equation is always true. To solve an equation, different operations are performed on either side of the equation according to the question provided. The main idea is to isolate the unknown to the left and all the known values to the right.

Lets consider different forms of solving an algebraic equation requiring only one step. Let C and D be considered as a representative of a real number and let x be a variable.

Form 1: $x + C = D$

Here, subtract C from both sides
$x + C - C = D - C$
$x = D – C$
Hence, if a number is added on one side containing the variable of the equation, then it is subtracted from both the sides.

Example:

$x + 2 = 1$

Solution:

$x + 2 = 1$
Subtract 2 from both the sides
$x + 2 - 2 = 1 - 2$
$x = 1 – 2$
$x = -1$

Form 2: $x - C = D$

Here, add C on both the sides.
$x - C + C = D + C$
$x = D + C$
Hence, if a number is subtracted from one side containing the variable of the equation, it is added on both the sides.

Example:

$x - 5 = 7$

Solution:

$x - 5 = 7$
Subtract 5 on both the sides
$x - 5 + 5 = 7 + 5$
$x = 7 + 5$
$x = 12$

Form 3
: $Cx = D$

Here, divide by C on both the sides.
$\frac{Cx}{C}$ = $\frac{D}{C}$
$x$ = $\frac{D}{C}$
Hence, if a number is multiplied on one side containing the variable of the equation, it is divided on both the sides.

Example:

$2x = 3$

Solution:

$2x = 3$
Subtract 2 on both the sides
$\frac{2x}{2}$ = $\frac{3}{2}$
$x$ = $\frac{3}{2}$

Form 4: $\frac{x}{C}$ = $D$

Here, multiply by C on both the sides.
$\frac{x}{C}$ $\times C$ = $D \times C$
$x = DC$
Hence, if a number is divided from one side containing the variable of the equation, it is multiplied on both the sides.

Example:

$\frac{x}{3}$ = $6$

Solution:

$\frac{x}{3}$ = $6$
Multiply by 3 on both the sides
$\frac{x}{3}$ $\times 3 = 6 \times 3$
$x = 18$

Transforming: Short-cut Method

When shifting from one side to other side, change the signs of each term. So, Form 1 and Form 2 can be written in the form of transforming in order to save even that one step.

Form 1: $x + C = D$
change $+C$ to $–C$ on the other side, hence $x = D – C$

Form 2
: $x - C = D$
change $-C$ to $+C$ on the other side, hence $x = D + C$

Example:

Solve $x + a – b + c = d$

Solution:

To solve x, transform + to – and – to +
So, $x + a – b + c = d$ changes to $x = d – a + b – c$
Hence solved.

In some equations, the unknown variable may fall on the right hand side. So, it’s better to exchange sides so as to bring it to the left hand side. In this process, the sign doesn’t change.

Example:

$1 = x + 6$
This is of Form 1. So, +6 changes to -6
Hence, $1 – 6 = x$
$-5 = x$
Exchanging the sides, we get $x = -5$

## Multi Step Algebraic Equations

Multi-step algebraic equations can be defined as an equation such that solving it will require more than two operations as steps. To solve it, the equation can be simplified by using the distributive property and combining like terms depending on the requirement.

Lets take an example of a simple multi-step algebraic equation and lets check the steps.
$2w + 3 = 6 + w$

The basic idea is to get all the variables on any of the side and all the numbers on the opposite side. Lets take the variables to the left and the numbers to the right. For that, subtract w from both sides
$2w + 3 - w = 6 + w - w$
$w + 3 = 6$
Subtract 3 from both the sides
$w + 3 - 3 = 6 - 3$
$w = 3$

In general, the steps followed for solving multi step algebraic equations are
• Isolate the variable and hence solution of equation can be obtained
• Use addition as inverse of subtraction and vice versa.
• Use division as inverse of multiplication and vice versa.
The numbers/constants used can be integers, fractions or decimals.

### Solving Multi-Step Algebraic Equations with Integers

To solve equations with integers, first we have to isolate the variables and then simplify the equation.

Example:

$3(u + 1) – (2u + 3) = 5$

First use distribution property and open the parenthesis,
$3u + 3 – 2u - 3 = 5$
Combine the like term
$3u – 2u + 3 – 3 = 5$
$u – 0 = 5$
$u = 5$

### Solving Multi-Step Algebraic Equations with Fractions

To solve equations with fraction, first of all simplify the equation without fraction.

Example:

$\frac{5x}{4}$ $+ 2 – 3x =6$

Distribute the terms by opening the parenthesis

$\frac{5x}{4}$
$+ 2 – 3x = 6$

Subtract 2 from both the sides

$\frac{5x}{4}$ $+ 2 – 3x - 2 = 6 - 2$

$\frac{5x}{4}$ $- 3x = 4$

Factor x on the left hand side

$\frac{(5 - 12)x}{4}$ = $4$
$\frac{-7x}{4}$ = $4$
Multiply by 4 on both the sides,
$-7x = 16$
Divide by (-7) on both the sides
$x$ = $\frac{16}{-7}$
$x$ = $\frac{-16}{7}$

### Solving Multi-Step Algebraic Equations with Decimals

To solve equations with decimals, first we have to isolate the variables and then simplify the equation.

Example 1:

$5(x – 2.3) – 2x = 5.3 + x$
Simplify as needed in the equation
$5(x – 2.3) – 2x = 5.3 + x$
Distribute the terms by opening the parenthesis
$5x – 11.5 – 2x = 5.3 + x$
Simplify all the like terms
$3x – 11.5 = 5.3 + x$
Subtract x from both the sides
$3x – 11.5 - x = 5.3 + x - x$
$2x – 11.5 = 5.3$
Add 11.5 on both the sides
$2x – 11.5 + 11.5 = 5.3 + 11.5$
$2x = 9.5$
Divide by 2 on both the sides
$\frac{2x}{2}$ = $\frac{16.8}{2}$
$x = 8.4$

Example 2:

Find three consecutive even integers whose sum is 24

Solution:

To solve lets assume the smallest even integer is $2x$.
Then, the next consecutive even integer = $2x + 2$
Next one = $(2x + 2) + 2 = 2x + 4$

Given sum of all of it is 24
So, $2x + (2x + 2) + (2x + 4) = 24$

Distribute the terms by opening the parenthesis
$2x + 2x + 2 + 2x + 4 = 24$

Simplify all the like terms
$6x + 6 = 24$

Subtract 6 from both the sides
$6x + 6 - 6 = 24 - 6$
$6x = 18$

Divide by 6 on both the sides
$\frac{6x}{6}$ = $\frac{18}{6}$
$x = 3$
So, the smallest number is $2 \times 3$ = $6$
Next is = $6 + 2$ = $8$
Last = $6 + 4$ = $10$

## Rules of Algebraic Equations

Algebra is a branch of mathematics that can help take a real situation and change it in the mathematical form. That is as an equation. As in real number, terms of algebra also follow a set of rules that helps in simplifying the equation and expression. Lets check out certain rules and their corresponding examples:

Basic Rules
Let x, y and z be any variable, then

Rule 1: Any term multiplied by 1 is that term itself.
$x \times 1 = x$
Here, 1 is called as the multiplicative identity.

Example:

$2 \times 1 = 2$
$5 \times 1 = 5$

Rule 2: Any term multiplied by -1 is the negative of that term.
$x \times (-1) = -x$

Example:
$8 \times (-1) = -8$
$2 \times (-1) = -2$

Rule 3: Negative of a negative term is negative.
$-(-x) = x$

Example:
$-(-3) = 3$

Rule 4: The product of a term and its reciprocal is 1.
$x \times ($ $\frac{1}{x}$ $) = 1$

Example:
$6 \times ($ $\frac{1}{6}$ $) = 1$

Rule 5: Sum of any term and zero is that term itself.
$x + 0 = x$
Here, 0 is called additive identity.

Example:
$6 + 0 = 6$

Rule 6: Sum of a positive term and its negative is zero
$x + (-x) = 0$
Hence, –x is the additive inverse of x.

Example:
$4 + (-4) = 0$

Rule 7: Combination of addition and subtraction sign is again a subtraction.
$x + (-y) = x + -y = x – y$

Example:
2 + (-3) = 2 – 3 which is -1

Rule 8: Combination of two subtraction sign is addition.
$x - (-y) = x + y$

Example:
$5 - (-3) = 5 + 3 = 8$

Rule 9: Multiplication of a negative and positive term is negative
$x \times (-y) = -xy$

Example:
$5 \times (-2) = -10$

Rule 10: Multiplication of two negative terms is a positive term
$(-x) \times (-y) = xy$

Example:
$(-2)(-5) = 10$

### Properties Concerning Algebraic Expressions

Addition:
• Commutative: Order of any set of terms doesn’t matter
$x + y = y + x$
Example:
$2 + 1 = 3 = 1 + 2$
• Associative: Consecutive numbers can be grouped accordingly
$(x + y) + z = x + (y + z)$
Example:
$(2 + 4) + 1 = 7 = 2 + (4 + 1)$

Multiplication
:
• Commutative: Order of any set of terms doesn’t matter
$x \times y = y \times x$
Example:
$4 \times 2 = 8 = 2 \times 4$
• Associative: Consecutive numbers can be grouped accordingly
$(x \times y) \times z = x \times (y \times z)$
Example:
$(3 \times 4) \times 2 = 24 = 3 \times (4 \times 2)$
• Distributive Property of Addition over Multiplication:
There are two types: Right and Left
Right: $(x + y)z = xz + yz$
Left: $z(x + y) = zx + zy$

Properties Using Equality

• In any equation, identical terms can be added on both sides, by which the equation remains unaltered.
So, if $x = y$, $x + z = y + z$
Example:
If $x = 7$, $x + 5 = 7 + 5$
$x + 5 = 12$
• In any equation, identical terms can be multiplied on both sides, by which the equation remains unaltered.
So, if $x = y$, $x \times z = y \times z$
Example:
If $u = 2$, $u \times 3 = 2 \times 3$
$3u = 6$
• In any equation, changing the sign on both the sides, makes the equation unaltered
So, if $x = y$, $-x = -y$
Example:
If $x = 5$, $-x = -5$

## Graphing Algebraic Equation

Graphing an algebraic equation is more of a visual way of looking at an equation so it is possible to understand its structure and properties. It is drawn with axes perpendicular to each other. In graphing a linear algebraic equation, two axes, x (horizontal axis) and y (vertical axis), perpendicular to each other, are taken. They intersect a point called as the origin. It is called the coordinate plane or the rectangular coordinate plane. Each point on the plane is represented by an ordered pair (a, b) where a is a point on the x axis and b is a point on the y axis.

There will be infinite points for an algebraic linear equation. But, its not possible to determine all of it. So, we use minimum points to get the whole equation graphed correctly. An algebraic linear equation can be mostly drawn in two ways.
• Using the slope of a line and the y intercept
• Using the T chart

### Using the Slope of a Line and the y-intercept

The standard form of writing any linear equation is y = (m) x + b
where m, being the x coefficient, is called the slope and b, being the constant, is called the y intercept.
The following procedure can to be undertaken:
• First the main point is to write the equation in the slope intercept form.
• Recognize the y intercept and plot it on the graph.
• Recognize the slope and using it determine another point on the xy plane.
• Join the two points and also extend it both to either side of the points to the required line
Example:

Graph $2y - x = 12$

First write it in the slope intercept form
$2y - x = 12$
Add x on both the side
So, $2y = x + 12$
Divide by 2 on both the sides
$y = ($ $\frac{1}{2}$ $) x + 6$
So, slope $m$ = $\frac{1}{2}$ and y intercept = (0, 6)

Plot the point (0, 6) on the graph.

From 6 go 1 units up, so it goes to 7 and 2 units right, so we end up at (2, 7)
Hence draw a line joining the two points.

### Using T chart

When graphing any algebraic linear equations, first form a chart, known as T- chart. It is called so by the shape it takes. It will have two columns. Left column contains x points and right column contains y points. You can choose any reasonable x values, plug it in the equation and solve for y. Grouping the x and corresponding y values makes a pair that can be used for plotting.

Any x values can be chosen such that y value exists.
For any straight line we need at least three points.

Draw a line joining all the three points.

Example :

Graph $y = x - 4$

For x values assume -1, 0, 1. Lets form T chart

 x y = x - 4 (x , y) -1 y = (-1) - 4 = -5 (-1, -5) 0 y = 0 - 4 = -4 (0, -4) 1 y = 1 - 4 = -3 (1, -3)
Plot this point on the graph and drawn the line joining the points.

## Types of Algebraic Equations

An algebraic equation can be defied by an equality separated by two algebraic expressions containing variables and constants. The set of algebraic equation can be generally grouped into five sets.

1) Polynomial Equation

A polynomial equation can be defined as an equation that contains variable terms whose power is a whole number. They are classified by the number of terms they contains. The classification is as follows:

Monomial:
It will only contain a single term
Example:
$2x = 0$

Binomial:
It will contain exactly two terms
Example:
$4x +7y = 0$

Trinomial:
It will contain exactly three terms
Example:
$2y + u – 4x = 0$

Polynomial equation can also be classified in terms of degrees
One degree (linear):
It is the simplest form of an algebraic equation. It is commonly known as a linear equation. It is of the form ax + b = c, where a, b, c are constants.
Example:
$3x + 1 = 0$

Two degree (quadratic):
An algebraic equation of degree 2 is called a quadratic equation. The general form is ax2 + bx + c = 0 where a $\neq$ 0 and a, b, c are constants.
Example:
5x2 - x +2

Cubical equation:
An algebraic equation of degree 3 is called a cubic equation. The general form is
ax3 + bx2 + cx + d = 0, where a $\neq$ 0 and a, b, c, d are constants.
Example:
3x3 + 2x2 + x + 3 = 0

2) Rational Equation
Any rational equation will be in the form y= p/q where both p and q are polynomials.
Example:
Y = $\frac{(2x – 5)}{(3x^{2} + x – 1)}$
Its very important the denominator mustn’t be zero.

3) Exponential Equation
Such equation contain variables in the exponent part.
Its form can be abx + c = 0.
If, the independent variable has a positive coefficient, it is exponential growth and if it has negative coefficient, then it is decay.
Example:
y = 3(2x-1) + 5

4) Logarithmic equation
Logarithm is the inverse of exponent.
So if y = ax is in the exponential equation then its logarithmic equation is x = loga y
The number ”e” is the most common logarithmic base and such a logarithm is called the natural logarithm.
Example:
Loga x +loga 4 = loga 3

5) Trigonometric Equations
It will contain trigonometric ratios. The peculiarity of a trigonometric function is that it will have periods so it will be repeat by itself after a particular point. Solving such an equation depends on certain rules and identities of trigonometry.
Example:
sinx + cos x = tan 2x