Q. 343 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}3x-8y=20\\ x+3y=1\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is  $\left(4,-1\right)$.

Step 1. Given information.

The system of linear equations, 

$\left\{\begin{array}{l}3x-8y=20\\ x+3y=1\end{array}\right.$

Step 2. Calculating the value of y .

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

Multiplying the second equation with $3$ we get

$3x+9y=3$

Then the equations will be,

$$\begin{gathered} 3x-8y=20 \\ 3x+9y=3 \end{gathered}$$

Subtracting the above equations then

$3x-8y=20-3x-9y=-3-17y=17y=-1$ 

Step 3. Calculating the value of x .

Substituting the value of $y$ in any equation.

$$\begin{gathered} 3x-8\left(-1\right)=20 \\ 3x+8=20 \\ 3x=12 \\ x=4 \end{gathered}$$ 

Step 4. Checking the result.

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

$$\begin{gathered} 3x-8y=20 \\ 3\left(4\right)-8(-1)=20 \\ 12+8=20 \\ 20=20 \end{gathered}$$

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

$$\begin{gathered} x+3y=1 \\ 4+3(-1)=1 \\ 4-3=1 \\ 1=1 \end{gathered}$$

Hence, this is true.