Question

1. ( x = 2y + 7 )

( 3x - 2y = 3 )

1. ( 4x - 2y = 14 )

( y = \frac{1}{2}x - 1 )

1. ( 2x = y - 10 )

( 2x + 4 = 2y )

1. ( 8x - \frac{1}{3}y = 0 )

Explanation:

Let's solve one of the systems of equations, say problem 3 and 4 (assuming they are a system: \( x = 2y + 7 \) and \( 3x - 2y = 3 \))

Step1: Substitute \( x \) from first equation into second

We have \( x = 2y + 7 \). Substitute this into \( 3x - 2y = 3 \). So replace \( x \) with \( 2y + 7 \) in the second equation: \( 3(2y + 7) - 2y = 3 \)

Step2: Simplify the equation

Expand \( 3(2y + 7) \): \( 6y + 21 - 2y = 3 \). Combine like terms: \( 4y + 21 = 3 \)

Step3: Solve for \( y \)

Subtract 21 from both sides: \( 4y = 3 - 21 = -18 \). Then divide by 4: \( y = \frac{-18}{4} = -\frac{9}{2} \)

Step4: Solve for \( x \)

Substitute \( y = -\frac{9}{2} \) into \( x = 2y + 7 \): \( x = 2(-\frac{9}{2}) + 7 = -9 + 7 = -2 \)

Answer:

\( x = -2 \), \( y = -\frac{9}{2} \)

If you want solutions for other problems (like 5,6 or 4), please clarify the system (which equations are paired) and I can solve them similarly.