Simultaneous equations are a system of equations carrying multiple variables.
Any set of equations that have to be solved together or simultaneously to get the value of unknown quantities which are true for each participating equations and will have unique values are better known as simultaneous equations.

Generally the solution of a system of equations is a particular value of all the variables that fulfills all the equations simultaneously(at the same time).
The standard form of system of linear equations in two variables x and y is:
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
Where a1,b1,c1,a2,b2 and c2 are all real numbers.


Equations which are carrying out more than one variable are generally known as simultaneous equations. There are three algebraic methods to solve the linear system of equations or simultaneous equations like
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0 are
(1) Elimination method
(2) Substitution method
(3) Cross - multiplication method.

The graphical method probably would not show the most accurate solution for the system of linear equations. Particularly, when the point corresponding to the solution having non-integral ordered-pairs like ( $\sqrt{5}$, $\sqrt{7}$ ) or ($\frac{2}{3}$, $\frac{6}{7}$)


To solve a linear system of equations by elimination method use the following steps.
Step 1: Convert both equations in the standard form Ax + By= C.
Step 2: Translate so that the coefficients of one pair of variable terms are reverse. Multiply one or both equations by accurate numbers so that the sum of the x-term or y-term coefficients is 0.
Step 3: Now, add the obtained equations to remove or eliminate a variable such that the sum contains only one variable.
Step 5: Substitute the solution from Step 3 into either of the original equations and work it out for the other variable.
Step 6: To check the answer, substitute the solution set in either or both the original equations.

Examples on the Elimination method


The following examples will explain how to solve simultaneous equations using the elimination method.

Example 1: Find the solution set using the elimination method.
x + y = -12
x - y = 2
Solution: Add the equations to eliminate y. then solve for x.
x + y = -12
x - y = 2
2x = -10
x = -5
Substitute -5 for x in the first equation, then solve for y.
-5 + y = -12
y = -7
Check by substituting -5 for x and -7 for y in both equations.
x + y = -12
$\Rightarrow$ -5 + (-7) = -12
$\Rightarrow$ -12 = -12
x - y = 2
$\Rightarrow$ -5 - (-7) = 2
$\Rightarrow$ -5 + 7 = 2
$\Rightarrow$ 2 = 2
Solution set: {(-5, -7)}

Example 2: Find the solution set using the elimination method.
x + 8y = -15
7x + 8y = 39
Solution: Multiply the first equation by -7, then add the result to the second equation to eliminate x.
solve for y.
-7(x + 8y) = -15(-7)
-7x - 56y = 105
7x + 8y = 39_
-48y = 144
y = -3
Substitute -3 for y in the first equation, then solve for x.
x + 8(-3) = -15
x - 24 = -15
x = 9
Check by substituting 9 for x and -3 for y in both equations.
x + 8y = -15
$\Rightarrow$ 9 + 8(-3) = -15
$\Rightarrow$ 9 - 24 =
$\Rightarrow$ -15 = -15
7x + 8y = 39
$\Rightarrow$ 7(9) + 8(-3) = 39
$\Rightarrow$ 63 - 24 = 39
$\Rightarrow$ 39 = 39
Solution set: {(9, -3)}

Example 3: Find the solution set using the elimination method.
$\frac{1}{2}$ x + $\frac{1}{2}$ y = -3
x - y = 8
Solution : Multiply equation 1 by 2, then add the result to the second equation to eliminate y.
2($\frac{1}{2}$ x + $\frac{1}{2}$ y) = -3(2)
x + y = -6
x - y = 8_
2x = 2
x = 1
Substitute 1 for x in the second equation and solve for y.
1 - y = 8
- y = 7
y = -7
Check by substituting 1for x and -7 for y in both the equations.
$\frac{1}{2}$ x + $\frac{1}{2}$ y = -3
$\Rightarrow$ $\frac{1}{2}$ (1) + $\frac{1}{2}$ (-7) = -3
$\Rightarrow$ $\frac{1}{2}$ + $\frac{-7}{2}$ = -3
$\Rightarrow$ $\frac{1 - 7}{2}$ = -3
$\Rightarrow$ $\frac{-6}{2}$ = -3
$\Rightarrow$ -3 = -3
x - y = 8
$\Rightarrow$ 1 - (-7) = 8
$\Rightarrow$ 1 + 7 = 8
$\Rightarrow$ 8 = 8
Solution set: {(1, -7)}

Example 4: Find the solution set using the elimination method.
-0.1x + 0.6y = 1.3
0.8x + 0.6y = -5
Solution : Multiply the second equation by -1, then add the result to the first equation to eliminate y.
-1(0.8x + 0.6y) = (-5)(-1)
-0.8x - 0.6y = 5
-0.1x + 0.6y = 1.3_
- 0.9x = 6.3
x = -7
Substitute -7 for x in the first equation and solve for y.
-0.1(-7) + 0.6y = 1.3
0.7 + 0.6y = 1.3
0.6y = 0.6
y = 1
Check by substituting -7 for x and 1 for y in both the equations.
-0.1x + 0.6y = 1.3
$\Rightarrow$ -0.1(-7) + 0.6(1) = 1.3
$\Rightarrow$ 0.7 + 0.6 = 1.3
$\Rightarrow$ 1.3 = 1.3
0.8x + 0.6y = -5
$\Rightarrow$ 0.8(-7) + 0.6(1) = -5
$\Rightarrow$ 0.8(-7) + 0.6(1) = -5
$\Rightarrow$ -5.6 + 0.6 = -5
$\Rightarrow$ -5 = -5
Solution set: {(-7, 1)}

Elimination Method Problems


The following practice problems help you to solve simultaneous equations using the elimination method. You can use the same format that we have shown in above given examples.

Problem 1: Find the solution using the elimination method.
x + 7y = -53
-5x + 6y = -22
Problem 2: Find the solution using the elimination method.
9x + 7y = -14
-7x + 2y = -4
Problem 3: Find the solution using the elimination method.
$\frac{3x}{8}$ - $\frac{3y}{5}$ = $\frac{33}{80}$
$\frac{4x}{7}$ + $\frac{4y}{5}$ = $\frac{37}{35}$
Problem 4: Find the solution using the elimination method.
-0.07x + 0.01y = -0.19
0.01x + 0.01y = -0.03

To solve a linear system of equations by substitution, use the following steps. Step 1: Solve one equation for either of the variables.
Step 2: Substitute that variable in the other equation and solve further.
Step 3: Now, substitute the result from step2 into either of the original equations to find the value of the other variable.
Step 4: To check the answer, substitute the solution set in either or both the original equations.

Examples on Substitution method


The following examples help you understand how to solve simultaneous equations using the substitution method.

Example 1: Find the solution set using the substitution method.
x - 6y = -18
2x - 7y = -16
Solution : Solve the first equation for x.
x - 6y = -18
x = 6y - 18
Now substitute 6y - 18 for x in the second equation, then solve for y.
2(6y - 18) - 7y = -16
12y -36 -7y = -16
5y - 36 = -16
5y = 20
y = 4
Substitute 4 for y into the first equation and solve for x.
x - 6(4) = -18
x - 24 = -18
x = 6
Check by substituting 6 for x and 4 for y in both equations.
x - 6y = -18
$\Rightarrow$ 6 - 6(4) = -18
$\Rightarrow$ 6 - 24 = -18
$\Rightarrow$ -18 = -18
2x - 7y = -16
$\Rightarrow$ 2(6) - 7(4) = -16
$\Rightarrow$ 12 - 28 = -16
$\Rightarrow$ -16 = -16
Solution set: {(6, 4)}

Example 2: Find the solution set using the substitution method.
x + 5y = 6
-2x + 6y = -12
Solution : Solve the first equation for x.
x + 5y = 6
x = 6 - 5y
Now substitute 6 - 5y for x in the second equation, then solve for y.
-2(6 - 5y) + 6y = -12
-12 + 10y + 6y = -12
-12 + 16y = -12
16y = 0
y = 0
Substitute 0 for y into the first equation and solve for x.
x + 5(0) = 6
x + 0 = 6
x = 6
Check by substituting 0 for x and 6 for y in both the equations.
x + 5y = 6
$\Rightarrow$ 6 + 5(0) = 6
$\Rightarrow$ 6 + 0 = 6
$\Rightarrow$ 6 = 6
-2x + 6y = -12
$\Rightarrow$ -2(6) + 6(0) = -12
$\Rightarrow$ -12 + 0 = -12
$\Rightarrow$ -12 = -12
Solution set: {(6, 0)}

Example 3: Find the solution set using the substitution method.
6x + 5y = 25
4x - 3y = -15
Solution: Solve the second equation for x.
4x - 3y = -15
4x = 3y - 15
x = $\frac{3y}{4}$ - $\frac{15}{4}$
Now substitute $\frac{3y}{4}$ - $\frac{15}{4}$ for x in the first equation and solve for x.
6($\frac{3y}{4}$ - $\frac{15}{4}$) + 5y = 25
$\frac{9y}{2}$ - $\frac{45}{2}$ + 5y = 25
9y - 45 + 10y = 50
19y - 45 = 50
19 y = 95
y = 5
Now substitute 5 for y in the second equation and solve for x.
4x - 3(5) = -15
4x - 15 = -15
4x = 0
x = 0
Check by substituting 5 for y and 0 for x in both the equations
6x + 5y = 25
$\Rightarrow$ 6(0) + 5(5) = 25
$\Rightarrow$ 0 + 25 = 25
$\Rightarrow$ 25 = 25
4x - 3y = -15
$\Rightarrow$ 4(0) - 3(5) = -15
$\Rightarrow$ 0 - 15 = -15
$\Rightarrow$ -15 = -15
Solution set: {(0, 5)}

Example 4: Find the solution set using the substitution method.
4x - 3y = 30 + x
4x = -(y + 2) + 3x
Solution : Solve the second equation for x.
4x = -(y + 2) + 3x
x = - (y + 2) = -y - 2
Substitute -y - 2 for x in the first equation, then solve for x.
4(-y - 2) - 3y = 30 + (-y - 2)
-4y - 8 - 3y = 30 - y -2
-7y - 8 = 28 - y
-6y - 8 = 28
-6y = 36
y = -6
Now substitute -6 for y in the second equation and solve for x.
4x = -(-6 + 2) + 3x
4x = -(-4) + 3x
4x = 4 + 3x
x = 4
Check by substituting 4 for x and -6 for y in both the equations
4x - 3y = 30 + x
$\Rightarrow$ 4(4) - 3(-6) = 30 + (4)
$\Rightarrow$ 16 + 18 = 34
$\Rightarrow$ 34 = 34
4x = -(y + 2) + 3x
$\Rightarrow$ 4(4) = -(-6 + 2) + 3(4)
$\Rightarrow$ 16 = -(-4) + 12
$\Rightarrow$ 16 = 4 + 12
$\Rightarrow$ 16 = 16
Solution set: {(4, -6)}

Substitution Method Problems


The following practice problems help you to solve simultaneous equations using the substitution method.

Problem 1: Find the solution set using the substitution method.
x + y = 6
x = y - 2
Problem 2: Find the solution set using the substitution method.
8x + 5y = -3
-2x + 3y = 5
Problem 3: Find the solution set using the substitution method.
$\frac{-x}{3}$ + $\frac{3y}{4}$ = $\frac{1}{2}$
$\frac{x}{6}$ + $\frac{y}{8}$ = $\frac{3}{4}$
Problems 4: Find the solution set using the substitution method.
-0.5x + 0.2y = 0.5
0.8x - 0.1y = 0.3

Let us consider the general form of the linear system of equations.
$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$
To solve for variables x and y by using cross-multiplication arrange the variables x and y and their respective coefficients a1, b1, a2 and b2 and constants c1 and c2 as
Cross - Multiplication Method
$\frac{x}{b_{1}c_{2} - b_{2} c_{1}}$ = $\frac{y}{c_{1}a_{2} - c_{2} a_{1}}$ = $\frac{1}{a_{1} b_{2} - a_{2} b_{1}}$
x = $\frac{b_{1} c_{2} - b_{2}c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
y = $\frac{c_{1} a_{2} - c_{2}a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

To solve a linear system of equations by cross-multiplication, use the following steps.
Step 1: Convert both the equations in the standard from Ax + By + C = 0
Step 2 : Observe the coefficients of the variables x and y and the constants.
Step 3: Apply the formula:
x = $\frac{b_{1} c_{2} - b_{2}c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
y = $\frac{c_{1} a_{2} - c_{2}a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

Examples on Cross Multiplication Method


The following examples help you to understand how to solve simultaneous equations using the cross multiplication method.

Example 1: Find the solution set using the cross-multiplication method.
x + y = 6
x - y = -2
Solution : Given equations are:
x + y = 6 $\Rightarrow$ x + y - 6 = 0
x - y = -2 $\Rightarrow$ x - y + 2 = 0
We can observe that a1 = 1, b1 = 1, c1 = -6, a2 = 1, b2 = -1, c2 = 2
Solution set: {(x, y)} is given by
x = $\frac{b_{1} c_{2} - b_{2}c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
x = $\frac{(1)(2) - (-1)(-6)}{(1)(-1) - (1)(1)}$
x = $\frac{2 - 6}{-1 - 1}$
x = $\frac{-4}{-2}$
x = 2
y = $\frac{c_{1} a_{2} - c_{2}a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
y = $\frac{(-6)(1) - (2)(1)}{(1)(-1) - (1)(1)}$
y = $\frac{-6 - 2}{-1 - 1}$
y = $\frac{-8}{-2}$
y = 4
Solution set: {(2, 4)}

Example 2: Find the solution set using the cross-multiplication method.
3x + 4y = 7
x + y = 2
Solution : Given equations are:
3x + 4y = 7 $\Rightarrow$ 3x + 4y - 7 = 0
x + y = 2 $\Rightarrow$ x + y - 2 = 0
We can observe that a1 = 3, b1 = 4, c1 = -7, a2 = 1, b2 = 1, c2 = -2
Solution set: {(x, y)} is given by
x = $\frac{b_{1} c_{2} - b_{2}c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
x = $\frac{(4)(-2) - (1)(-7)}{(3)(1) - (1)(4)}$
x = $\frac{-8 + 7}{3 - 4}$
x = $\frac{-1}{-1}$
x = 1
y = $\frac{c_{1} a_{2} - c_{2}a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$
y = $\frac{(-7)(1) - (-2)(3)}{(3)(1) - (1)(4)}$
y = $\frac{-7 + 6}{3 - 4}$
y = $\frac{-1}{-1}$
y = 1
Solution set: {(1, 1)}

Cross Multiplication Method Problems


The following practice problems help you to solve simultaneous equations using the cross multiplication method.

Problem 1: Find the solution set using the cross-multiplication method.
2x+3y= 4
-x+4y= 9
Problem 2: Find the solution set using the cross-multiplication method.
x + 2y = 7
2x +3y = 5
Problem 3: Find the solution set using the cross-multiplication method.
3x+2y =12
4x+3y = 17
Problem 4: Find the solution set using the cross-multiplication method.
2x + y = 5
x + 2y = 1

The standard form of the system of linear equations in two variables x and y
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
Where a1,b1,c1,a2,b2 and c2 are all real numbers.

These pair of linear equations when plotted on a graph define two lines and only one of the below three possibilities can happen

a) Two lines intersect at one point
If $\frac{a_{1}}{a_{2}}$ $\neq$ $\frac{b_{1}}{b_{2}}$, then the system of linear equations
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0 has a unique solution that is exactly one solution and the equations are said to be a consistent pair of linear equations.

b) The two lines coincident with each other, that is they lie exactly on top of each other.
If $\frac{a_{1}}{a_{2}}$ = $\frac{b_{1}}{b_{2}}$ = $\frac{c_{1}}{c_{2}}$, then the system of linear equations
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0 has an infinite solution that is exactly one solution and the equations are said to be consistent pair of linear equations.

c) Two lines are parallel to each other, that is they will not intersect
If $\frac{a_{1}}{a_{2}}$ = $\frac{b_{1}}{b_{2}}$ $\neq$ $\frac{c_{1}}{c_{2}}$, then the system of linear equations
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0 has no solution and the equations are said to be inconsistent pair of linear equations.
Types of Linear System

Examples on Graphical Method of Solution


The following examples help you understand how to solve simultaneous equations using the graphical method.

Example 1: Solve by graphing. The system given below has a unique solution (consistent system).
x - y = 2
x + y = 4
Solution :
For equation 1
x - y = 2
Subtract x on both sides
x - y - x = 2 - x
-y = 2 - x
Multiply with (-1) on both sides
(-1) $\times$ (-y) = (-1)(2 - x)
y = -2 + x
y = x - 2
Now let us assign some values and find the corresponding y-values to plot on the graph
x
0 2
3
y = x - 2
y = x - 2
= 0 - 2
= -2
y = x - 2
= 2 - 2
= 0
y = x - 2
= 3 - 2
= 1

For equation 2
x + y = 4
Subtract x on both sides
x + y - x = 4 - x
y = 4 - x
Let us assign some values and find the corresponding y-values to plot on the graph

x 0
2 3
y = 4 - x y = 4 - x
= 4 - 0
= 4
y = 4 - x
= 4 - 2
= 2
y = 4 - x
= 4 - 3
= 1

Now let us plot these ordered pairs
Examples on Graphical Method of Solution
The two linear graphs intersect at the point (3, 1).

Example 2: Solve by graphing. The system given below has a no solution (inconsistent system).
2x + y = 5
2x + y = 8
Solution :
For equation 1
2x + y = 5
Subtract 2x on both sides
2x + y - 2x = 5 - 2x
y = 5 - 2x

Now let us assign some values and find the corresponding y-values to plot on the graph

x
0 2
4
y = 5 - 2x
y = 5 - 2x
= 5 - 2(0)
= 5
y = 5 - 2x
= 5 - 2(2)
= 1
y = 5 - 2x
= 5 - 2(4)
= -3

For equation 2
2x + y = 8
Subtract 2x on both sides
2x + y - 2x = 8 - 2x
y = 8 - 2x
Let us assign some values and find the corresponding y-values to plot on the graph

x
0
2
4
y = 8 - 2x
y = 8 - 2x
= 8 - 2(0)
= 8
y = 8 - 2x
= 8 - 2(2)
= 4
y = 8 - 2x
= 8 - 2(4)
= 0

Now let us plot these ordered pairs
Examples on Graphical Method of Solution
From the above we can see that two lines are parallel to each other. So the system has no solution.
The general linear equation is 
$ax + by + c = 0$ , $a^{2} + b^{2} \neq 0$

A pair of linear equations is called a system of equations.

$a_{1}x + b_{1}y + c_{1} = 0$, $(a_{1})^{2} + (b_{1})^{2} \neq 0$
$a_{2}x + b_{2}y + c_{2} = 0$, $(a_{2})^{2} + (b_{2})^{2} \neq 0$

Here, the solution exists if, 
$a_{1} b_{2} - a_{2} b_{1} \neq 0$

The ordered pair (x,y) which satisfies both the equations is called a solution of the system of the linear equations. This solution can be obtained in four different ways. Substitution method is one of the four ways.

As the name indicates, in the substitution method, we need to substitute the value of one variable derived from one equation into the other equation.

Example 1:

$2x + y = 5$
$x + 2y = 1$

Solution:

$2x + y = 5$ ------------ (1)
$x + 2y = 1$ ------------ (2)

From equation (1), we have
$2x + y = 5$
$y = 5 - 2x$ -------------- (3)

Substituting $y = 5 - 2x$ in equation (2), we get

$x + 2(5 - 2x) = 1$
$x + 10 - 4x = 1$
$-3x = 1 - 10$
$-3x = -9$
$x = 3$

Now, substituting $x = 3$ in equation (3), we get

$y = 5 - 2x$
     = $5 - 2(3)$
     = $5 - 6$
     = $-1$

Hence (3, -1) is the solution of the given system of equations.
The general linear equation is 
$ax + by + c = 0$ , $a^{2} + b^{2} \neq 0$

A pair of linear equations is called a system of equations.

$a_{1}x + b_{1}y + c_{1} = 0$, $(a_{1})^{2} + (b_{1})^{2} \neq 0$
$a_{2}x + b_{2}y + c_{2} = 0$, $(a_{2})^{2} + (b_{2})^{2} \neq 0$

Here, the solution exists if, 
$a_{1} b_{2} - a_{2} b_{1} \neq 0$

The ordered pair (x,y) which satisfies both the equations is called a solution of the system of the linear equations. This solution can be obtained in four different ways. Elimination method is one of the four ways.

In the elimination method, we first make the coefficient of either x or y as the additive inverse of each other.

Example 1:

$x + 2y = 7$
$2x + 3y = 5$

Solution:

$x + 2y = 7$ ------------- (1)
$2x + 3y = 5$ ----------- (2)

Here, we make the coefficients of y as the additive inverses of each other by multiplying equation (1) by 3 and equation (2) by -2.

$3x + 6y = 21$
$-4x - 6y = -10$

Adding these equations, we get 

$-x = 11$ or $x = -11$

Substituting $x = -11$ in (1), we get 

$-11 + 2y = 7$
$2y = 7 + 11$
$2y = 18$
$y = 9$

Hence the solution is (-11, 9)