Question

question #5 a student is trying to solve the systems of equations. the students work is shown. did the student make a mistake? if so, in which step was the mistake made? what was the mistake? 3x² + 4y - 7x = 20 2x - y + 4 = 0 step 1: 2x - y + 4 = 0 y = 2x + 4 step 2: 3x² + 4(2x + 4) - 7x - 20 = 0 3x² + 8x + 16 - 7x - 20 3x² + x - 4 = 0 step 3: (3x + 4)(x - 1) = 0 x = -4/3 x = 1 step 4: x = -4/3 x = 1 y = 2(-4/3)+4 y = 2(1)+4 y = 4/3 y = 6 step 5: (-4/3,4/3) (1,6) step 3; the student has the wrong factors. the factors should be (3x - 4)(x + 1). step 5;the student has the wrong solutions. the only answer which works is (1,9). step 1;the student solved for the linear equation incorrectly. it should be y = 2x - 4. no errors were made. the work is correct.

Explanation:

Step1: Check Step 1

Starting with $2x - y+4 = 0$, when we solve for $y$, we add $y$ to both sides and subtract 0 from both sides to get $y=2x + 4$. This step is correct.

Step2: Check Step 2

Substitute $y = 2x + 4$ into $3x^{2}+4y-7x = 20$. We have $3x^{2}+4(2x + 4)-7x-20=0$. Expand: $3x^{2}+8x + 16-7x-20=3x^{2}+x - 4=0$. This step is correct.

Step3: Check Step 3

Factor $3x^{2}+x - 4=0$. Using the formula for factoring $ax^{2}+bx + c$ where $a = 3$, $b = 1$, $c=-4$ and $ac=-12$. We need two numbers that multiply to - 12 and add up to 1. The numbers are 4 and - 3. So $3x^{2}+x - 4=3x^{2}+4x-3x - 4=x(3x + 4)-(3x + 4)=(3x + 4)(x - 1)=0$. Then $x=-\frac{4}{3}$ or $x = 1$. This step is correct.

Step4: Check Step 4

When $x=-\frac{4}{3}$, $y=2(-\frac{4}{3})+4=\frac{-8 + 12}{3}=\frac{4}{3}$. When $x = 1$, $y=2(1)+4=6$. This step is correct.

Step5: Check Step 5

The solution pairs are $(x,y)=(-\frac{4}{3},\frac{4}{3})$ and $(1,6)$. This step is correct.

Answer:

No errors were made. The work is correct.