MAST Grade 8 Systems of Equations. Practice it free.

Both equations have slope $2$ but different $y$-intercepts ($3$ and $-1$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = -x + 5$ and $y = 3x - 3$. The slopes $-1$ and $3$ are different, so the lines intersect at exactly one point.

Both equations have slope $-3$ but different $y$-intercepts ($2$ and $5$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = -2x + 7$ and $y = x - 2$. The slopes $-2$ and $1$ are different, so the lines intersect at exactly one point.

Both equations have slope $4$ but different $y$-intercepts ($-1$ and $3$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = \dfrac{5}{3}x - \dfrac{4}{3}$ and $y = -2x + 7$. The slopes $\dfrac{5}{3}$ and $-2$ are different, so the lines intersect at exactly one point.

Both equations have slope $-1$ but different $y$-intercepts ($6$ and $-2$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = \dfrac{1}{2}x - \dfrac{3}{2}$ and $y = \dfrac{1}{2}x - 2$. The slopes are the same, but the $y$-intercepts differ, so the lines are parallel and the system has zero solutions.