Question

1. solve the system of equations below. \

$$\begin{cases} 3x + 2y = 8 \\\\ 5x - 2y = 12 \\end{cases}$$

fill in the ordered pair that represents the solution (\square, \square)

Explanation:

Step1: Eliminate \( y \) by adding the two equations

We have the system:
\[

$$\begin{cases} 3x + 2y = 8 \\ 5x - 2y = 12 \end{cases}$$

\]
Adding the left - hand sides and the right - hand sides of the two equations:
\((3x + 2y)+(5x - 2y)=8 + 12\)
Simplify the left - hand side: \(3x+5x+2y - 2y=8x\), and the right - hand side: \(20\). So we get the equation \(8x=20\).

Step2: Solve for \( x \)

To solve for \( x \), divide both sides of the equation \(8x = 20\) by \(8\):
\(x=\frac{20}{8}=\frac{5}{2}=2.5\)

Step3: Substitute \( x=\frac{5}{2} \) into one of the original equations to solve for \( y \)

Let's substitute \( x = \frac{5}{2}\) into the first equation \(3x+2y = 8\).
\(3\times\frac{5}{2}+2y=8\)
\(\frac{15}{2}+2y = 8\)
Subtract \(\frac{15}{2}\) from both sides:
\(2y=8-\frac{15}{2}\)
We know that \(8=\frac{16}{2}\), so \(2y=\frac{16}{2}-\frac{15}{2}=\frac{1}{2}\)
Divide both sides by \(2\): \(y=\frac{1}{2}\div2=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4} = 0.25\)

Answer:

\((\frac{5}{2},\frac{1}{4})\) or \((2.5,0.25)\)