CalculatorUser

1. Multiply second equation by 2

\(\frac{5}{3}\)x-2y=-2

-2x+2y=4

2. Use elimination to cancel the y's

-2x+\(\frac{5}{3}\)x=2

3. Now its your turn

Solve the equation for x, plug it back in one of the equations, then you get your answer.

(I want you to do it yourself so you learn)

ohhhhh i want to facepalm myself now. thank you!

Sorry i didn't see your reply

No this problem has ONE solution, the ordered pair (5,-3) is the ONLY solution. So ONE solution

y=x-8

2x=3y=1

2x+3(x-8)=1

2x+3x-24=1

5x-24=1

5x=25

\(\boxed{x=5}\), then plug it in to one of the equations

y=x-8

y=5-8

\(\boxed{y=-3}\)

I'm not sure if that's possible, if you are looking to for the answer in terms of height and area. Then it is\(\boxed{h=\frac{2a}{21}}\) .

You must have an area to find out the height or else it isn't possible.

Thank you so much!

3400 - 38 = 3362 (not freshman)

38 - 35 = 3 (freshman opposed)

3362 - 15 = 3347 (non-freshman opposed)

3347 + 3 = 3350

\(\boxed{3350}\) Students are opposed

1. Note the equation \(\frac{Y_2-Y_1}{X_2-X_1}\)

2. Substitute \(\frac{(30)-(16)}{(5)-(2)}\)

3. Solve \(\frac{14}{3}\)

4.  Note the equation \(y=mx+b\)

We already have \(m\), which is \(\frac{14}{3}\)

So \(y=\frac{14}{3}x+b\)

5. Substitute a coordinate \((x,y)\), from the problem

Take \((2,16)\) and substitute it in to get: \(16=\frac{14}{3}*(2)+b\)

Solving we get \(b=\frac{20}{3}\)

6. Put \(b\) and \(m\) together to make the equation \(\boxed{y=\frac{14}{3}x+\frac{20}{3}}\)