Question

t is the solution to the system of \

$$\begin{cases} 4x + 2y = 20 \\\\ 3x + 2y = 10 \\end{cases}$$

\bigcirc (10, -10) \bigcirc (5, 0) \bigcirc (20, 10) (0, 10)

Explanation:

Step1: Subtract the two equations

We have the system:

$$\begin{cases}4x + 2y = 20\\3x + 2y = 10\end{cases}$$

Subtract the second equation from the first: $(4x + 2y)-(3x + 2y)=20 - 10$.
Simplify: $4x+2y - 3x - 2y=10$, so $x = 10$.

Step2: Substitute x into an equation

Substitute $x = 10$ into the second equation $3x + 2y = 10$:
$3(10)+2y = 10$.
Simplify: $30+2y = 10$.
Subtract 30 from both sides: $2y=10 - 30=-20$.
Divide by 2: $y=-10$.

Answer:

(10, -10)