The elimination method is a way of solving a simultaneous equation in which one variable is eliminated so that the other variable can be solved easily.

The steps for solving simultaneous equation by elimination are:

1. Arrange the equation so that all the variables are on one side of the equation and the constant on the other side of the equations.

2. Modify one of the equations so that both equations share common variables that can be eliminated.

3. Depending on the case, add or subtract both equations so that one variable is eliminated.

4. Solve the other variable not eliminated.

5. Substitute the value of a variable into one of the equations above.

And remember to check your answer after solving by plugging the value of x and y into the equation.

๐This post is an extension of our post on simultaneous equation๐

Let's practice with these four examples

__Example 1__Using elimination method, solve the simultaneous equation $2x-y=11$, $x+2y=-7$

**Solution:**

This equation is already in the normal form of simulteanous equation,

$2x-y=11$....eqn 1

$x+2y=-7$....eqn 2

To eliminate, you first modify one of the equations, let's modify Eqn 2 by multiplying it by 2

Multiplying eqn 2 by 2

$2(x+2y=-7)$

$2x+4y=-14$

Now let's substract the two equations

$\begin{matrix} 2x-y=11\\ -(2x+4y=-14)\\ \hline 0-5y=25 \end{matrix}$

$-5y=25$

$y=-5$

To solve for $x$, we would Insertthe value of $y$ in eqn 1

$2x-(-5)=11$

$2x+5=11$

$2x=11-5$

$2x=6$

$\frac{2x}{2}=\frac{6}{2}$

$x=3$

__Example 2__Solve the simultaneous equation $3x+4y=0$, $x=2y-5$

**Solution:**

Unlike the previous one, this equation is not arranged in the usual form of the simultaneous equation, hence we would rearrange it like this:

$3x+4y=0$....eqn 1

$x-2y=-5$....eqn 2

Now, let's modify Eqn 2 by multiplying by 3

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

$3x-6y=-15$

Substracting the equations

$\begin{matrix} 3x+4y=0\\ -(3x-6y=-15)\\ \hline 0+10y=15 \end{matrix}$

$10y=15$

Divide both sides by 10

$\frac{10y}{10}=\frac{15}{10}$

$y=\frac{3}{2}$

To solve for x, we will Insert the value of $y$ in eqn 1

$3x+4(\frac{3}{2})=0$

$3x+\frac{12}{2}=0$

Taking the L.C.M

$\frac{6x+12}{2}=0$

Cross multiply

$6x+12=0\times2$

$6x+12=0$

$6x=-12$

Divide bothh sides by 6

$\frac{6x}{6}=\frac{-12}{6}$

$x=-2$

**Example 3**Solve the equation using simulteanous equation $x+3y=-13$, $2x-9y=4$

**Solution:**

$x+3y=-13$......eqn 1

$2x-9y=4$..... eqn 2

Let multiply eqn 1 by 2

$2(x+3y=-13)$

$2x+6y=-26$

Substracting the equations

$\begin{matrix} 2x+6y=-26\\ -(2x-9y=4)\\ \hline 0+15y=-30 \end{matrix}$

$15y=-30$

$\frac{15y}{15}=\frac{-30}{15}$

$y=-2$

Inserting the value of x in eqn 2

$2x-9(-2)=4$

$2x+18=4$

$2x=4-18$

$2x=-14$

$x=-7$

__Related posts__

**Example 4**By elimination, find the solution to $x-2y=1$, $x+2y=9$

**Solution:**

Unlike the previous ones, without modifying any equation, $x$ can be eliminated. So, we subtract the two equations

$\begin{matrix} x-2y=1\\ -(x+2y=9)\\ \hline 0-4y=-8 \end{matrix}$

$-4y=-8$

$\frac{-4y}{-4}=\frac{-8}{-4}$

$y=2$

Inserting the value of y in eqn 1

$x-2(2)=1$

$x-4=1$

$x=1+4$

$x=5$

That will be all for now. We shall conclude our series on simultaneous equations in this next post.

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