Question

final exam - requires respondus lockdown brow
started dec 10 at 6:56pm
quiz instructions
question 39
solve the system by the elimination method or substitution method
x + 2y = -4
8x + 1y = 58
○ {(7, -5)}
○ {(8, -6)}
○ {(7, -5)}
○ {(-8, -5)}

Explanation:

Step1: Use substitution method

From the first equation \(x + 2y=-4\), we can express \(x\) as \(x=-4 - 2y\).

Step2: Substitute into second equation

Substitute \(x=-4 - 2y\) into the second equation \(8x + y = 58\). We get \(8(-4 - 2y)+y = 58\).

Step3: Simplify the equation

Expand the left - hand side: \(-32-16y + y=58\). Combine like terms: \(-32-15y = 58\).
Add 32 to both sides: \(-15y=58 + 32=90\).
Divide both sides by - 15: \(y=\frac{90}{-15}=-6\). Wait, this seems wrong. Wait, let's re - check. Wait the second equation is \(8x + y = 58\)? Wait maybe I misread the problem. Wait the user's problem: the first equation is \(x + 2y=-4\), the second equation: maybe it's \(8x + y = 58\)? Wait no, let's do it again. Wait, if we use elimination. Multiply the first equation by 8: \(8x+16y=-32\). The second equation is \(8x + y = 58\). Subtract the second equation from the first (after multiplying the first by 8): \((8x + 16y)-(8x + y)=-32 - 58\). \(15y=-90\), so \(y=-6\). Then substitute \(y = - 6\) into \(x+2y=-4\), \(x+2(-6)=-4\), \(x-12=-4\), \(x = 8\). Wait but the options: one of the options is \((8,-6)\). Wait let's check the options again. The options are: \((7,5)\), \((8,-6)\), \((7,-5)\), \((-8,-5)\). Let's check \((8,-6)\): for \(x + 2y\): \(8+2\times(-6)=8 - 12=-4\) (correct). For \(8x + y\): \(8\times8+(-6)=64 - 6 = 58\) (correct). So the correct solution is \((8,-6)\).

Answer:

\((8,-6)\) (corresponding to the option with \((8,-6)\))