Question

1. \

$$\begin{cases} y = 2x - 1 \\\\ 3x - 4y = -40 \\end{cases}$$

Explanation:

Step1: Substitute \( y = -2x - 1 \) into \( 3x - 4y = -40 \)

We have the system of equations:

$$
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$$

Substitute \( y = -2x - 1 \) into the second equation:
\( 3x - 4(-2x - 1) = -40 \)

Step2: Simplify and solve for \( x \)

Expand the left - hand side:
\( 3x + 8x + 4 = -40 \)
Combine like terms:
\( 11x+4=-40 \)
Subtract 4 from both sides:
\( 11x=-40 - 4=-44 \)
Divide both sides by 11:
\( x=\frac{-44}{11}=-4 \)

Step3: Substitute \( x = - 4 \) into \( y=-2x - 1 \) to find \( y \)

\( y=-2\times(-4)-1 \)
\( y = 8 - 1=7 \)

Answer:

The solution of the system of equations is \( x=-4,y = 7 \) (the point of intersection of the two lines is \((-4,7)\))