Question

4
andre solves an equation, but when he checks his answer he notices that his solution is incorrect. he knows he made an error, but he can’t find it. where is andre’s error and what is the solution to the equation?
-2(3x - 5) = 4(x + 3) + 8
-6x + 10 = 4x + 12 + 8
-6x + 10 = 4x + 20
10 = -2x + 20
-10 = -2x
5 = x

5
from unit 3, lesson 13
choose the equation that has solutions (5, 7) and (8, 13).
a. 3x - y = 8
b. y = x + 2
c. y - x = 5
d. y = 2x - 3

6
from unit 3, lesson 9
a length of ribbon is cut into two pieces to use in a craft project. the graph shows the length (in feet) of the second piece, x, for each length of the first piece, y.

Explanation:

Question 4

Step1: Analyze the step \(10 = -2x + 20\)

To isolate \(x\), we should subtract \(20\) from both sides or add \(6x\) to both sides. Let's check the step where Andre goes from \(-6x + 10 = 4x + 20\) to \(10 = -2x + 20\). He should add \(6x\) to both sides: \(-6x+6x + 10=4x + 6x+20\), which gives \(10 = 10x + 20\), not \(10=-2x + 20\). So the error is in the step where he combines like terms (adding \(6x\) to \(4x\) incorrectly, he subtracted \(6x\) from \(4x\) instead of adding).

Step2: Solve the equation correctly

Start with \(-2(3x - 5)=4(x + 3)+8\)
Expand both sides: \(-6x + 10 = 4x+12 + 8\)
Simplify right side: \(-6x + 10 = 4x+20\)
Add \(6x\) to both sides: \(10 = 10x+20\)
Subtract \(20\) from both sides: \(10 - 20=10x\)
Simplify: \(-10 = 10x\)
Divide by \(10\): \(x=-1\)

To determine which equation has solutions \((5,7)\) and \((8,13)\), we substitute \(x\) and \(y\) values into each equation.

• Option A: \(3x - y = 8\)

Substitute \((5,7)\): \(3(5)-7=15 - 7 = 8\), which works. Substitute \((8,13)\): \(3(8)-13=24 - 13 = 11
eq8\), so A is incorrect.

• Option B: \(y = x + 2\)

Substitute \((5,7)\): \(7=5 + 2=7\), works. Substitute \((8,13)\): \(13=8 + 2=10
eq13\), so B is incorrect.

• Option C: \(y - x = 5\)

Substitute \((5,7)\): \(7 - 5 = 2
eq5\), so C is incorrect.

• Option D: \(y = 2x - 3\)

Substitute \((5,7)\): \(2(5)-3=10 - 3 = 7\), works. Substitute \((8,13)\): \(2(8)-3=16 - 3 = 13\), works.

Answer:

Andre's error is in the step \( -6x + 10 = 4x + 20\) to \(10 = -2x + 20\) (incorrect combination of like terms). The solution is \(x = -1\).

Question 5