Question

In the diagram l || m. Find the value(s) of x.m<1(x^2-7x)m<7=(-x+7)

Answer 1

Given:

m∠1 = (x² - 7x)

m∠7 = (-x + 7)

m∠1 and m∠7 are alternate exterior angles, and alternate exterior angles are congruent.

Therefore,

m∠1 = m∠7

(x² - 7x) = (-x + 7)

Let's solve for the values of x:

Move all the terms to the left hand side and equate to zero

$x^2-7x+x-7=0$

Let's factorize:

$x(x-7)+1(x-7)$

Now, the factors are:

(x + 1) and (x - 7)

Equate each factor to zero and solve for x:

x + 1 = 0

x = 0 - 1

x = -1

x - 7 = 0

x = 0 + 7

x = 7

Therefore, the values of x are:

-1 and 7

ANSWER:

x = -1 and 7