Q 55 - Intermediate Algebra

Question

In the following exercises, solve the systems of equations by elimination.

$\left\{\begin{array}{l}2x+9y=-4\\ 3x+13y=-7\end{array}\right.$

Step-by-Step Solution

Answer

The solution is$\left(-11,2\right)$.

Step 1. Given information

The linear equation given are:

$$\begin{gathered} 2x+9y=-4 \\ 3x+13y=-7 \end{gathered}$$

Step 2. Calculating the value of y

Now, to solve the equation using elimination we have to do some changes in the equations so that the $x$ 's term will be cancelled.

1) multiply the first equation by $3$

$3\left(2x+9y=-4\right)$

$6x+27y=-12$

2) multiply the second equation by $-2$

$-2\left(3x+13y=-7\right)$

$-6x-26y=14$

Now the new equations are 

$$\begin{gathered} 6x+27y=-12 \\ -6x-26y=14 \end{gathered}$$

now, just add the above equations

$6x+27y=-12-6x-26y=14y=2$

the value of $y$ is $2$

Step 3. Calculating the value of x

As we have the value of y now. we can substitute it in any of the equation to calculate the value of $x$.

let us take the first equation 

$2x+9y=-4$

put $y$=2

$2x=-22$

$x=-11$

the value of $x$:  $-11$

Step 4. Checking the solution

Checking the solution by putting the value of $x,y$ in the equation, we get

$$\begin{gathered} 2x+9y=4 \\ 2(-11)+9\left(2\right)=4 \\ -22+18=4 \\ 4=4 \\ 3x+13y=-7 \\ 3(-11)+13\left(2\right)=-7 \\ -33+26=-7 \\ -7=-7 \\ LHS=RHS \end{gathered}$$

This is true, hence the solution is correct.