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 (in feet) of the second piece, x, for each length of the first piece, y.

Explanation:

Question 4

Step1: Analyze Andre's steps

Andre's equation: $-2(3x - 5)=4(x + 3)+8$
First step: $-6x + 10 = 4x + 12 + 8$ (distributed correctly)
Second step: $-6x + 10 = 4x + 20$ (combined like terms correctly)
Third step: $10 = -2x + 20$ (Here, he should add $6x$ to both sides: $-6x + 6x+10 = 4x + 6x+20$ → $10 = 10x + 20$. His error is in the sign when moving $-6x$ to the right. He subtracted $6x$ instead of adding, leading to $10=-2x + 20$ (incorrect).

Step2: Solve the equation correctly

Start from $-6x + 10 = 4x + 20$
Add $6x$ to both sides: $10 = 10x + 20$
Subtract $20$: $10 - 20 = 10x$ → $-10 = 10x$
Divide by $10$: $x = -1$

To check which equation has solutions $(5,7)$ and $(8,13)$, substitute $x$ and $y$ into each option.

• Option A: $3x - y = 8$

For $(5,7)$: $3(5)-7 = 15 - 7 = 8$ (works). For $(8,13)$: $3(8)-13 = 24 - 13 = 11
eq 8$ (fails).

• Option B: $y = x + 2$

For $(5,7)$: $7 = 5 + 2 = 7$ (works). For $(8,13)$: $13 = 8 + 2 = 10
eq 13$ (fails).

• Option C: $y - x = 5$

For $(5,7)$: $7 - 5 = 2
eq 5$ (fails).

• Option D: $y = 2x - 3$

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

Answer:

Andre's error is in the step $10 = -2x + 20$ (incorrectly moved $-6x$; should add $6x$ to both sides). The correct solution is $x = -1$.

Question 5