Question

directions: if l || m, solve for x and y. 9. (9x + 25)° (13x - 19)° (17y + 5)° 10. (3x - 29)° (8y + 17)° (6x - 7)° 11. (7x - 23)° 49° (3x)° (11y - 1)° 12. (7y - 20)° (5x - 38)° (3x - 4)°

Explanation:

Step1: Use property of alternate - interior angles for \(x\) in problem 9

Since \(l\parallel m\), the alternate - interior angles \((9x + 25)^{\circ}\) and \((13x-19)^{\circ}\) are equal. So, \(9x + 25=13x-19\).

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Step2: Use property of corresponding angles for \(y\) in problem 9

The corresponding angles \((13x - 19)^{\circ}\) and \((17y + 5)^{\circ}\) are equal. Substitute \(x = 11\) into \((13x-19)\): \(13\times11-19=143 - 19=124\). Then \(17y+5 = 124\).

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Step3: Use property of alternate - exterior angles for \(x\) in problem 10

Since \(l\parallel m\), the alternate - exterior angles \((3x-29)^{\circ}\) and \((6x - 7)^{\circ}\) are equal. So, \(3x-29=6x - 7\).

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Step4: Use property of corresponding angles for \(y\) in problem 10

The corresponding angles \((3x - 29)^{\circ}\) and \((8y + 17)^{\circ}\) are equal. Substitute \(x=-\frac{22}{3}\) into \(3x-29\): \(3\times(-\frac{22}{3})-29=-22-29=-51\). Then \(8y + 17=-51\).

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Step5: Use vertical - angle property for \(x\) in problem 11

The vertical angles \((3x)^{\circ}\) and \(49^{\circ}\) are equal. So, \(3x = 49\), and \(x=\frac{49}{3}\).

Step6: Use property of corresponding angles for \(y\) in problem 11

The corresponding angles \((7x - 23)^{\circ}\) and \((11y - 1)^{\circ}\) are equal. Substitute \(x = \frac{49}{3}\) into \(7x-23\): \(7\times\frac{49}{3}-23=\frac{343}{3}-23=\frac{343 - 69}{3}=\frac{274}{3}\). Then \(11y-1=\frac{274}{3}\).

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Step7: Use property of same - side interior angles for \(x\) in problem 12

The same - side interior angles \((5x-38)^{\circ}\) and \((3x - 4)^{\circ}\) are supplementary (since \(l\parallel m\)), so \((5x-38)+(3x - 4)=180\).

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Step8: Use property of same - side interior angles for \(y\) in problem 12

The same - side interior angles \((7y-20)^{\circ}\) and \((5x - 38)^{\circ}\) are supplementary. Substitute \(x=\frac{111}{4}\) into \(5x-38\): \(5\times\frac{111}{4}-38=\frac{555}{4}-38=\frac{555 - 152}{4}=\frac{403}{4}\). Then \((7y-20)+\frac{403}{4}=180\).

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Answer:

Problem 9: \(x = 11\), \(y = 7\)
Problem 10: \(x=-\frac{22}{3}\), \(y =-\frac{17}{2}\)
Problem 11: \(x=\frac{49}{3}\), \(y=\frac{277}{33}\)
Problem 12: \(x=\frac{111}{4}\), \(y=\frac{397}{28}\)