Question

solve each system of equations by graphing

1. \\(\

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

\\)

\\(\

$$\begin{cases} y = x - 1 \\\\ x + 4y = 16 \\end{cases}$$

\\)

Explanation:

Step1: Solve the first system \(

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

\)
Set the two equations equal: \(\frac{2}{3}x-1=-x + 4\)
Add \(x\) to both sides: \(\frac{2}{3}x+x-1=4\)
Combine like terms: \(\frac{2x + 3x}{3}-1=4\), \(\frac{5x}{3}-1 = 4\)
Add 1 to both sides: \(\frac{5x}{3}=5\)
Multiply both sides by \(\frac{3}{5}\): \(x = 3\)
Substitute \(x = 3\) into \(y=-x + 4\): \(y=-3 + 4=1\)

Step2: Solve the second system \(

$$\begin{cases}y=x - 1\\x + 4y=16\end{cases}$$

\)
Substitute \(y=x - 1\) into \(x + 4y=16\): \(x+4(x - 1)=16\)
Expand: \(x+4x-4 = 16\)
Combine like terms: \(5x-4=16\)
Add 4 to both sides: \(5x=20\)
Divide by 5: \(x = 4\)
Substitute \(x = 4\) into \(y=x - 1\): \(y=4 - 1=3\)

Answer:

First system solution: \(x = 3,y = 1\)
Second system solution: \(x = 4,y = 3\)