# SIMULTEANOUS EQUATION BY SUBSTITUTION

A simultaneous equation is one that consists of two-equation made up of two variables.

To obtain the particular solution to a simultaneous equation, we must find a numerical value for each variable in the equations that will satisfy all equations simultaneously.

Consider the following example
$3x-4y=7$
$x-2y=5$

In the above equation, the solution would be $x=-3$ and $y=-4$

want to check?
$3(-3)-4(-4)=7$  true
$-3-2(-4)=5$ true

There are two ways of solving simultaneous, which are substitution and elimination methods.

We shall be focusing on the substitution method in this blog post.

To solve by substitution,
1. You make one of the variables the subject of the formula
2. The next step is to insert the derived equation into the other equation.
3. Then, you solve for the other variable
4. After that, then you back-substitute the value of the variable into one of the equations.

Ok, Let's practice with these four examples

Example 1
Solve the simultaneous equation
$4x-2y=8$, $6x+2y=22$

$4x-2y=8$....eqn 1
$6x+2y=22$....eqn 2

Making x the subject of the formula in Eqn 1
$4x-2y=8$
$4x=8+2y$

Divide through by 4
$\frac{4x}{4}=\frac{8}{4}+\frac{2y}{4}$
$x=2+\frac{y}{2}$.....eqn 3

Substituting eqn 3 in eqn 2
$6(2+\frac{y}{2})+2y=22$
$12+\frac{6y}{2}+2y=22$

Taking the L.C.M of the right hand side
$\frac{24+6y+4y}{2}=22$
$\frac{24+10y}{2}=22$

Cross multiply
$24+10y=44$
$10y=44-24$
$10y=20$

Divide both sides by 10
$\frac{10y}{10}=\frac{20}{10}$
$y=2$

Inserting the value of x in eqn 3
$x=2+\frac{y}{2}$
$x=2+\frac{2}{2}$

Taking the L.C.M
$x=\frac{4+2}{2}$
$x=3$

Example 2
Solve the simultaneous equation
$x+y=6$, $2x+3y=14$

$x+y=6$.....eqn 1
$2x+3y=14$.....eqn 2

Making x the subject of the formula in eqn 1
$x+y=6$
$x=6-y$.......eqn 3

Substituting eqn 3 into eqn 2
$2(6-y)+3y=14$
$12-2y+3y=14$
$12+y=14$
$y=14-12$
$y=2$

Inserting the value of y in eqn 3
$x=6-2$
$x=4$

Example 3
Using substitution method, solve this simulteanous equation $x-2y=1$, $x=9-2y$

$x-2y=1$.......eqn 1
$x=9-2y$.......eqn 2

Unlike the previous examples,  $x$ is already the subject of the formula in this example. So, we insert eqn 1 in Eqn 2.
$(9-2y)-2y=1$
$9-2y-2y=1$
$9-4y=1$
$-4y=1-9$
$-4y=-8$

Divide both side by $-4$
$\frac{-4y}{-4}=\frac{-8}{-4}$
$y=2$

Inserting the value of $y$ in eqn 2
$x=9-2(2)$
$x=9-4$
$x=5$

Related post
Example 4
Solve the simultaneous equation, $4x-3y=1$, $2x+4y=17$.

$4x-3y=1$.......eqn 1
$2x+4y=17$.....eqn 2

Making x the subject of the formula in Eqn 1
$4x-3y=1$
$4x=1+3y$

Divide through by 4
$\frac{4x}{4}=\frac{1}{4}+\frac{3y}{4}$
$x=\frac{1+3y}{4}$.......eqn 3

Inserting the eqn 3 in eqn 2
$2(\frac{1+3y}{4})+4y=17$
$\frac{2+6y}{4}+4y=17$

Taking the L.C.M
$\frac{2+6y+16y}{4}=17$
$\frac{2+22y}{4}=17$

Cross multiply
$2+22y=68$
$22y=68-2$
$22y=66$

Divide through by 22
$\frac{22y}{22}=\frac{66}{22}$
$y=3$

Inserting the value of y in eqn 3
$x=\frac{1+3y}{4}$
$x=\frac{1+3(3)}{4}$
$x=\frac{10}{4}$
$x=\frac{5}{2}$

There you have it. In our next post, you will learn how to use the elimination method to solve simultaneous equations.

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