Q. 344 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}9x+4y=2\\ 5x+3y=5\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is  $\left(-2,5\right)$.

Step 1. Given information.

The system of linear equations, 

$\left\{\begin{array}{l}9x+4y=2\\ 5x+3y=5\end{array}\right.$

Step 2. Calculating the value of y

Now, we will make some changes in the equations to eliminate $x$ terms.

Multiplying the first equation with $5$ we get

$$\begin{gathered} 5\left(9x+4y=2\right) \\ 45x+20y=10 \end{gathered}$$

Multiplying the second equation with $9$ we get

 $$\begin{gathered} 9\left(5x+3y=5\right) \\ 45x+27y=45 \end{gathered}$$

Then the equations will be,

$$\begin{gathered} 45x+20y=10 \\ 45x+27y=45 \end{gathered}$$

Subtracting the above equations then 

$45x+20y=10-45x-27y=-45-7y=-35y=5$

Step 3. Calculating the value of x .

Substituting the value of $y$ in any equation.

$$\begin{gathered} 9x+4\left(5\right)=2 \\ 9x+20=2 \\ 9x=-18 \\ x=-2 \end{gathered}$$

Step 4. Checking the result.

Substituting the value of $x$ and $y$ in the first equation then

$$\begin{gathered} 9x+4y=2 \\ 9(-2)+4\left(5\right)=2 \\ -18+20=2 \\ 2=2 \end{gathered}$$

Substituting the value of $x$ and $y$ in the second equation then

$$\begin{gathered} 5x+3y=5 \\ 5(-2)+3\left(5\right)=5 \\ -10+15=5 \\ 5=5 \end{gathered}$$

Hence, this is true.