Question

for problems 14 and 15, parallel lines ( ell ) and ( m ) are cut by a transversal ( n ). write and solve an equation to find the value for ( x ).
14.
15.

Explanation:

Problem 14

Step1: Set up the equation

Since \( \ell\parallel m\) and \(n\) is a transversal, the angles \(2(x + 20)^{\circ}\) and \(3(x-5)^{\circ}\) are congruent (alternate - interior angles). So, \(2(x + 20)=3(x - 5)\).

Step2: Expand the equation

Using the distributive property \(a(b + c)=ab+ac\), we get \(2x+40 = 3x-15\).

Step3: Solve for \(x\)

Subtract \(2x\) from both sides: \(2x - 2x+40=3x-2x - 15\), which simplifies to \(40=x - 15\). Then add \(15\) to both sides: \(x=40 + 15\).

Problem 15

Step1: Set up the equation

Since \( \ell\parallel m\) and \(n\) is a transversal, the angles \((x - 18)^{\circ}\) and \(2(x - 53)^{\circ}\) are congruent (corresponding angles). So, \(x-18=2(x - 53)\).

Step2: Expand the equation

Using the distributive property \(a(b - c)=ab-ac\), we get \(x-18=2x-106\).

Step3: Solve for \(x\)

Subtract \(x\) from both sides: \(x-x - 18=2x-x-106\), which simplifies to \(-18=x - 106\). Then add \(106\) to both sides: \(x=-18 + 106\).

Answer:

• For problem 14: \(x = 55\)
• For problem 15: \(x = 88\)