Question

solve each system of equations using an algebraic method.

1. \\(\

$$\begin{cases} 3x + 5y = 15 \\\\ 3x - 2y = 6 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} y = -7x + 12 \\\\ y = 5x - 36 \\end{cases}$$

\\)

Explanation:

Problem 18:

Step 1: Subtract the second equation from the first to eliminate \(x\)

We have the system:
\[

$$\begin{cases} 3x + 5y = 15 \\ 3x - 2y = 6 \end{cases}$$

\]
Subtract the second equation from the first: \((3x + 5y)-(3x - 2y)=15 - 6\)
Simplify: \(3x+5y - 3x + 2y=9\) which gives \(7y = 9\)

Step 2: Solve for \(y\)

From \(7y=9\), divide both sides by 7: \(y=\frac{9}{7}\)

Step 3: Substitute \(y\) into one of the original equations to find \(x\)

Substitute \(y = \frac{9}{7}\) into \(3x-2y = 6\):
\(3x-2\times\frac{9}{7}=6\)
\(3x-\frac{18}{7}=6\)
Add \(\frac{18}{7}\) to both sides: \(3x=6+\frac{18}{7}=\frac{42 + 18}{7}=\frac{60}{7}\)
Divide by 3: \(x=\frac{60}{7}\div3=\frac{60}{7}\times\frac{1}{3}=\frac{20}{7}\)

Problem 19:

Step 1: Set the two expressions for \(y\) equal to each other

We have the system:
\[

$$\begin{cases} y=-7x + 12\\ y=5x-36 \end{cases}$$

\]
Since both equal \(y\), set them equal: \(-7x + 12=5x-36\)

Step 2: Solve for \(x\)

Add \(7x\) to both sides: \(12=12x-36\)
Add 36 to both sides: \(48 = 12x\)
Divide by 12: \(x = 4\)

Step 3: Substitute \(x\) into one of the equations for \(y\)

Substitute \(x = 4\) into \(y=5x-36\):
\(y=5\times4-36=20 - 36=-16\)

Answer:

1. \(x=\boldsymbol{\frac{20}{7}}\), \(y=\boldsymbol{\frac{9}{7}}\)
2. \(x=\boldsymbol{4}\), \(y=\boldsymbol{-16}\)