KAP Grade 8 Expressions and Equations. Practice it free.

Both equations have slope $6$ but different $y$-intercepts ($1$ and $-4$), so the lines are parallel and the system has no solution.

Rewrite in slope-intercept form: $y = -\dfrac{1}{3}x + \dfrac{4}{3}$ and $y = -\dfrac{1}{3}x + \dfrac{4}{3}$. The slopes and $y$-intercepts match, so the equations describe the same line and the system has infinitely many solutions.

Multiply every term by $2$ to clear the fraction: $x + 8 = 2x - 2$. Subtract $x$ from both sides: $8 = x - 2$. Add $2$ to both sides: $x = 10\textsf{.}$

Subtract $1$ from both sides: $4y - 1 = 3x$. Divide both sides by $3$: $x = \dfrac{4y - 1}{3}$.

The Quotient of Powers Property says that powers with the same nonzero base are divided by subtracting the exponents, written $\dfrac{b^{m}}{b^{n}} = b^{m-n}\textsf{.}$ Here both the numerator and the denominator have base $b$, so $\dfrac{b^{8}}{b^{4}} = b^{8-4} = b^{4}\textsf{.}$

In $y = mx + b$, the $y$-intercept is the point $(0, b)$. Here $b = 4$, so the $y$-intercept is $(0, 4)\textsf{.}$

The first equation simplifies to $y = -5x + 6\textsf{,}$ and the second simplifies to $y = -\dfrac{10}{m}x + \dfrac{11}{m}\textsf{.}$ When $m = 2\textsf{,}$ $-\dfrac{10}{m} = -5\textsf{,}$ so the lines have the same slope, and since $6$ does not equal $\dfrac{11}{2}\textsf{,}$ the lines are parallel with different $y$-intercepts, so the system has no solution.

Rewrite in slope-intercept form: $y = -\dfrac{3}{4}x + \dfrac{5}{2}$ and $y = \dfrac{1}{2}x - \dfrac{1}{2}$. The slopes $-\dfrac{3}{4}$ and $\dfrac{1}{2}$ are different, so the lines intersect at exactly one point.