Question: 15. $y = \frac{1}{3}x + 2$ $y = \frac{2}{3}x + 5$ 17. $9x + 3y = -3$ $2x - y = -4$ 19. $x -…

15. $y = \frac{1}{3}x + 2$

$y = \frac{2}{3}x + 5$

17. $9x + 3y = -3$

$2x - y = -4$

19. $x - 4y = -4$

$-3x - 4y = 12$

Solution

To solve each system of equations: 15. \[ \begin{cases} y = \frac{1}{3}x + 2 & \\ y = \frac{2}{3}x + 5 & \end{cases} \] Set the equations equal to each other since both are equal to \(y\). \(\frac{1}{3}x + 2 = \frac{2}{3}x + 5\) Subtract \(\frac{1}{3}x\) from both sides: \(2 = \frac{1}{3}x + 5\) Subtract 5 from both sides: \(2 - 5 = \frac{1}{3}x\) \(-3 = \frac{1}{3}x\) Multiply both sides by 3: \(x = -9\) Substitute \(x = -9\) back into the first equation: \(y = \frac{1}{3}(-9) + 2\) \(y = -3 + 2\) \(y = -1\) Solution: \(x = -9, y = -1\) 17. \[ \begin{cases} 9x + 3y = -3 & \\ 2x - y = -4 & \end{cases} \] Multiply the second equation by 3 to eliminate \(y\): \(6x - 3y = -12\) Add the two equations: \[ \begin{align*} 9x + 3y &= -3 \\ 6x - 3y &= -12 \\ \hline 15x &= -15 \\ \end{align*} \] Divide by 15: \(x = -1\) Substitute \(x = -1\) back into the second equation: \(2(-1) - y = -4\) \(-2 - y = -4\) Add 2 to both sides: \(-y = -2\) Multiply by -1: \(y = 2\) Solution: \(x = -1, y = 2\) 19. \[ \begin{cases} x - 4y = -4 & \\ -3x - 4y = 12 & \end{cases} \] Multiply the first equation by 3 to align the \(x\) terms: \(3x - 12y = -12\) Add the equations: \[ \begin{align*} 3x - 12y &= -12 \\ -3x - 4y &= 12 \\ \hline -16y &= 0 \\ \end{align*} \] Divide by -16: \(y = 0\) Substitute \(y = 0\) back into the first equation: \(x - 4(0) = -4\) \(x = -4\) Solution: \(x = -4, y = 0\)