To solve for $\displaystyle y$ for $\displaystyle x=3$, we have to plug $\displaystyle 3$ into the $\displaystyle x$ variable of the equation and solve for $\displaystyle y$:

$\displaystyle 5(3)+2y=13$

$\displaystyle 15+2y=13$

$\displaystyle 2y=13-15$

$\displaystyle 2y=-2$

$\displaystyle y=-1$

To solve for $\displaystyle x$ for $\displaystyle y=1$, we have to plug 1 into the $\displaystyle y$ variable of the equation and solve for $\displaystyle x$:

$\displaystyle 3y=4x+1$

$\displaystyle 3(1)=4x+1$

$\displaystyle 3=4x+1$

$\displaystyle 3-1=4x$

$\displaystyle 2=4x$

$\displaystyle x=\frac{1}{2}$

To solve for $\displaystyle y$ for $\displaystyle x=5$, we have to plug $\displaystyle 5$ into the $\displaystyle x$ variable of the equation and solve for $\displaystyle y$:

$\displaystyle 14y-10x=20$

$\displaystyle 14y-10(5)=20$

$\displaystyle 14y-50=20$

$\displaystyle 14y=20+50$

$\displaystyle 14y=70$

$\displaystyle y=5$

First, use the points to find the equation of JK:

Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

$\displaystyle m=\frac{y'-y}{x'-x}$

Plug in and calculate:

$\displaystyle \small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}$

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

$\displaystyle \small 5=\frac{1}{2}*4+b$

$\displaystyle \small b=5-2=3$

So our answer is: 

$\displaystyle \small y=\frac{x}{2}+3$

To find y, we need to plug in 16 for x and solve:

$\displaystyle \small \small y=\frac{16}{2}+3=8+3=11$

To find out if (-2,5) is on f(x), simply plug the point into f(x)

$\displaystyle \small f(x)=4x+13$

Becomes,

$\displaystyle \small \small 5=4*-2+13=-8+13=5$

So yes, it does!

One way to solve this one is by plugging in each of the answer choices and eliminating any that don't work out. Begin with our original g(x)

$\displaystyle {}g(x)=4x^2-8$

If we plug in 3 we get

$\displaystyle {}g(x)=4(3)^2-8=4(9)-8=36-8=28$, 

So our point is (3,28).

Since the slope and a point on the line are given, we can use the point-slope formula:

$\displaystyle y-y_{1}=m(x-x_{1})$

$\displaystyle m=1\ and\ (1,4)$

$\displaystyle y-4=1(x-1)$

$\displaystyle y-4=x-1$

$\displaystyle y=x-1+4$

$\displaystyle y=x+3$

Since the slope and a point on the line are given, we can use the point-slope formula:

$\displaystyle y-y_{1}=m(x-x_{1})$

$\displaystyle m=0\ and\ (5,2)$

$\displaystyle y-2=0(x-5)$

$\displaystyle y-2=0$

$\displaystyle y=2$

First find the slope of the equation.

$\displaystyle m =\frac{rise}{run} =\frac{7 + 2}{1-4} = \frac{9}{-3}=-3$

Now plug in one of the two points to form an equation.  Here we use (4, -2), but either point will produce the same answer.

$\displaystyle y-(-2)=-3(x-4)$

$\displaystyle y + 2=-3x + 12$

$\displaystyle y=-3x+10$