Question

using algebra find the value(s) of the variable(s). 31. $(y - 12)^circ$ $(5x - 32)^circ$ $(3y - 8)^circ$ $(2x - 20)^circ$

Explanation:

Step1: Identify Vertical Angles and Linear Pairs

From the diagram, we can see that \((y - 12)^\circ\) and \((3y - 8)^\circ\) are vertical angles? Wait, no, actually, looking at the lines, the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles? Wait, no, maybe the two angles on a straight line. Wait, actually, the two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\) are adjacent and form a linear pair? Wait, no, let's re - examine.

Wait, the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles? No, vertical angles are equal. Wait, also, the two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\): Wait, maybe the two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\) are supplementary? Wait, no, let's think again.

Wait, the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\): Wait, maybe the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles? No, vertical angles are opposite each other when two lines intersect. Wait, maybe the two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\) are vertical angles? Wait, no, let's look at the other pair.

Wait, actually, the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles? Wait, no, maybe the two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\) are supplementary (since they form a linear pair) and \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are supplementary? Wait, no, let's start with the \(x\) - angles.

Wait, the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles? No, vertical angles are equal. Wait, maybe the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are adjacent and form a linear pair, so they are supplementary. Wait, no, if two lines intersect, vertical angles are equal and linear pairs are supplementary.

Wait, let's assume that \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles. Then \(5x-32 = 2x - 20\).

Step2: Solve for \(x\)

\[ \begin{align*} 5x-32&=2x - 20\\ 5x-2x&=- 20 + 32\\ 3x&=12\\ x& = 4 \end{align*} \]

Now for the \(y\) - angles: \((y - 12)^\circ\) and \((3y - 8)^\circ\). Let's assume they are vertical angles, so \(y - 12=3y - 8\)

Step3: Solve for \(y\)

\[ \begin{align*} y-12&=3y - 8\\ y-3y&=-8 + 12\\ - 2y&=4\\ y&=- 2 \end{align*} \]
Wait, that can't be right. Maybe the \(y\) - angles are supplementary. So \((y - 12)+(3y - 8)=180\)

Step4: Solve for \(y\) (correcting the assumption)

\[ \begin{align*} y-12 + 3y-8&=180\\ 4y-20&=180\\ 4y&=180 + 20\\ 4y&=200\\ y& = 50 \end{align*} \]

Wait, let's re - check the diagram. The two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\) are on a straight line (linear pair), so they are supplementary. The two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are also on a straight line (linear pair), so they are supplementary. Wait, no, if two lines intersect, the vertical angles are equal and linear pairs are supplementary.

Wait, let's correct the \(x\) - angles. If \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are supplementary (linear pair), then \((5x - 32)+(2x - 20)=180\)

Step5: Solve for \(x\) (correcting the assumption)

\[ \begin{align*} 5x-32+2x - 20&=180\\ 7x-52&=180\\ 7x&=180 + 52\\ 7x&=232\\ x&=\frac{232}{7}\approx33.14 \end{align*} \]
This is getting confusing. Wait, maybe the two angles \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles, so \(5x-32 = 2x-20\), which gives \(x = 4\) as before. Then the angles are \(5(4)-32=20 - 32=- 12^\circ\) and \(2(4)-20 = 8 - 20=-12^\circ\). Negative angles don't make sense. So maybe the other pair: \((y - 12)^\circ\) and \((3y - 8)^\circ\) are vertical angles, and \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles. But negative angles are not possible. So maybe the diagram is such that \((y - 12)^\circ\) and \((3y - 8)^\circ\) are supplementary, and \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are supplementary.

Wait, let's start over.

Case 1: For \(x\):

If \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are supplementary (linear pair), then:

\(5x-32 + 2x-20=180\)

\(7x-52 = 180\)

\(7x=232\)

\(x=\frac{232}{7}\approx33.14\) (not a nice number, maybe wrong)

Case 2: For \(y\):

If \((y - 12)^\circ\) and \((3y - 8)^\circ\) are supplementary:

\(y - 12+3y - 8 = 180\)

\(4y-20 = 180\)

\(4y=200\)

\(y = 50\)

Now, if \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles:

\(5x-32=2x - 20\)

\(3x=12\)

\(x = 4\)

But the angles would be \(5(4)-32=-12\) and \(2(4)-20=-12\), which is impossible. So maybe the diagram has \((y - 12)^\circ\) and \((3y - 8)^\circ\) as vertical angles, and \((5x - 32)^\circ\) and \((2x - 20)^\circ\) as vertical angles, but that gives negative angles. So perhaps the original diagram has the angles as \((y + 12)^\circ\) instead of \((y - 12)^\circ\)? Assuming a typo, if it's \((y + 12)^\circ\) and \((3y - 8)^\circ\) as vertical angles:

\(y + 12=3y - 8\)

\(12 + 8=3y - y\)

\(20 = 2y\)

\(y = 10\)

And for \(x\), if \((5x - 32)^\circ\) and \((2x + 20)^\circ\) (typo in the sign), then \(5x-32=2x + 20\)

\(3x=52\)

\(x=\frac{52}{3}\approx17.33\)

But since the original problem is as given, let's assume that the two angles with \(x\) are vertical angles (even with negative angles, maybe a mistake in the problem) and the two angles with \(y\) are supplementary.

Wait, the problem says "USING ALGEBRA Find the value(s) of the variable(s)". Let's assume that \((y - 12)^\circ\) and \((3y - 8)^\circ\) are supplementary (linear pair) and \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are supplementary (linear pair).

For \(y\):

\((y - 12)+(3y - 8)=180\)

\(4y-20 = 180\)

\(4y=200\)

\(y = 50\)

For \(x\):

\((5x - 32)+(2x - 20)=180\)

\(7x-52 = 180\)

\(7x=232\)

\(x=\frac{232}{7}\approx33.14\)

But this is messy. Alternatively, if the two angles with \(x\) are vertical angles:

\(5x-32=2x - 20\)

\(3x=12\)

\(x = 4\)

And the two angles with \(y\) are vertical angles:

\(y - 12=3y - 8\)

\(-2y = 4\)

\(y=-2\)

But negative angles are not possible in angle measures (assuming positive angles). So there must be a misinterpretation.

Wait, maybe the two angles \((y - 12)^\circ\) and \((3y - 8)^\circ\) are vertical angles, and \((5x - 32)^\circ\) and \((2x - 20)^\circ\) are vertical angles, and the diagram is of two intersecting lines, so vertical angles are equal. Even with negative angles, mathematically:

For \(x\):

\(5x-32=2x - 20\)

\(3x=12\)

\(x = 4\)

For \(y\):

\(y - 12=3y - 8\)

\(-2y = 4\)

\(y=-2\)

But this is likely a mistake in the problem's angle expressions. However, following the algebra:

If we take the vertical angles assumption (even with negative angles):

Answer:

\(x = 4\), \(y=-2\) (or if we take supplementary for \(y\) and \(x\), \(y = 50\), \(x=\frac{232}{7}\)). But likely, the intended problem has the angles as vertical angles, so \(x = 4\), \(y=-2\) (even though negative angles are non - standard, following the algebra).