Q 39. - Intermediate Algebra

Question

Solve the system of equations by substitution

$\left\{\begin{array}{l}4x+y=10\\ x-2y=-20\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations is,

$$\begin{gathered} x=0 \\ y=10 \end{gathered}$$

Step 1. Given the information

The system of equations is,

$\left\{\begin{array}{l}4x+y=10\\ x-2y=-20\end{array}\right.$

Step 2. Solve the first equation for y .

Solving the equation $4x+y=10$ for $y$

$4x+y=10$

$y=10-4x$

Step 3. Substituting the value of y in the equation.

Substituting $y=10-4x$ in the equation

$x-2y=-20$

$x-2(10-4x)=-20$

$x-20+8x=-20$

$9x-20=-20$

$9x=-20+20$

$9x=0$

$x=0÷9$

$x=0$

Step 4. Substituting the value of x in the equation.

Substituting $x=0$ in the equation

$y=10-4x$

$y=10-4\left(0\right)$

$y=10-0$

$y=10$

Step 5. Checking the solution for the equation 4 x + y = 10

Substituting $$\begin{gathered} x=0 \\ y=10 \end{gathered}$$in the equation

$4x+y=10$

$4\left(0\right)+10=10$

$0+10=10$

$10=10$

This is true.

Step 6. Checking the solution for the equation x - 2 y = - 20

Substituting $$\begin{gathered} x=0 \\ y=10 \end{gathered}$$in the equation

$x-2y=-20$

$0-2\left(10\right)=-20$

$0-20=-20$

$-20=-20$

This is true.