I'm not sure about the second image.
But the first one is telling you to translate x, 3 units left, and translate y, 4 units up.
(-6, 0) will turn to (-9, 4)
I subtracted 3 from -6 and added 4 to 0.
Answer:
x = -3
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
Step-by-step explanation:
<u>Step 1: Define equation</u>
5(x + 4) = -2(-4 - x) + 3
<u>Step 2: Solve for </u><em><u>x</u></em>
- Distribute: 5x + 20 = 8 + 2x + 3
- Combine like terms: 5x + 20 = 2x + 11
- Subtract 2x on both sides: 3x + 20 = 11
- Subtract 20 on both sides: 3x = -9
- Divide 3 on both sides: x = -3
<u>Step 3: Check</u>
<em>Plug in x to verify it's a solution.</em>
- Substitute: 5(-3 + 4) = -2(-4 - -3) + 3
- Simplify: 5(-3 + 4) = -2(-4 + 3) + 3
- Add: 5(1) = -2(-1) + 3
- Multiply: 5 = 2 + 3
- Add: 5 = 5
Here we see that 5 does indeed equal 5. ∴ x = -3 is a solution of the equation.
And we have our final answer!
The equation is y=-5x + 20, x being the hours that go by after 5pm.
Check the picture below. So let's use those two points on the line.

bearing in mind that the standard form is also a general form.
standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
Answer:
the answer is 58 dear because there he couldn't figure out what he meant