Grade 9 Math: Linear Relations

The factor $-2$ reflects in the $x$-axis and stretches vertically by 2. The $(x+3)$ shifts left 3. The $+1$ shifts the graph up 1.

Inverse variation for fixed distance uses $t=\text{distance}/\text{speed}$. With 60 km, $t=60/v$. The other options are not inverse or not $t$ as a function of $v$.

Set $2x+1=-x+7$ so $3x=6$ and $x=2$. Substitute: $y=2(2)+1=5$. The solution is $(2,5)$.

Substitute each $x$ into $y=2x+1$. For $(3,7)$, $y=2(3)+1=7$, so it works. The others give $y=9$, $y=-1$, and $y=1$, which do not match their listed $y$-values.

Use $m=\dfrac{y_2-y_1}{x_2-x_1}=(1-5)/(6-2)=-4/4=-1$. So the slope is $-1$.

The fixed fee is the $y$-intercept $15$. The cost increases by $0.10 per minute, the slope. So $C=0.10m+15$.

Perpendicular slopes are negative reciprocals: $m_1=-0.5$ gives $m_2=-1/(-0.5)=2$. Only $y=2x-4$ has slope $2$.

Shaded below means $y$-values are less than the line. A solid boundary means the line is included, so use $\le$.

A linear relation has the form $y=mx+b$. Only $y=3x-5$ fits this form. The others are quadratic, reciprocal, or absolute value, so they are non-linear.

For the $x$-intercept, set $y=0$: $3x=12\Rightarrow x=4$. For the $y$-intercept, set $x=0$: $2y=12\Rightarrow y=6$.

Slope $m=(10-2)/(1-(-3))=8/4=2$. Using $2=2(-3)+b$ gives $b=8$. So the equation is $y=2x+8$.