Question

step 2: $13x = -26$
step 3: $x = -2$
step 4: $3(-2)+2y = -5$
$-6 + 2y = -5$
$2y = 1$
$y = \frac{1}{2}$
step 5: the solution is $\left(-2, \frac{1}{2}\
ight)$.
is hugo’s solution to the system of equations correct? explain why or why not.

• yes, hugo’s solution is correct.
• no, hugo should have gotten positive 2 in step three.
• no, hugo forgot to multiply both sides of the first equation in step one.
• no, hugo should have subtracted the equations instead of adding them in step 2

Explanation:

Step1: Verify Step 3 calculation

$\frac{-26}{13} = -2$, so $x=-2$ is correct.

Step2: Verify Step 4 substitution

Substitute $x=-2$ into $3x+2y=-5$:
$3(-2)+2y=-5 \implies -6+2y=-5$

Step3: Solve for y

$2y=-5+6=1 \implies y=\frac{1}{2}$

Step4: Check solution validity

Substitute $(-2,\frac{1}{2})$ back into the original system (assuming the first equation was $10x+8y=-26$ to get Step2's $13x=-26$ when combined with $3x+2y=-5$ scaled appropriately):
For $3x+2y$: $3(-2)+2(\frac{1}{2})=-6+1=-5$, which matches.
For the other equation: $10(-2)+8(\frac{1}{2})=-20+4=-16$, and if the combined step gave $13x=-26$, the scaling was correct.

Answer:

Yes, Hugo's solution is correct.