Question

Find x satisfying each of the following equations:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer 1

(i)

If x > 3 then = x − 3

⇒ x − 3 = 2

⇒ x = 2 + 3 = 5

If x < 3 then = − (x − 3) = 3 − x

⇒ 3 − x = 2

⇒ x = 3 − 2 = 1

∴ x = 1 and 5

(ii)

If x > − 2 then = x + 2

⇒ x + 2 = 1

⇒ x = 1 − 2 = − 1

If x < − 2 then = − (x + 2) = −x − 2

⇒ −x − 2 = 1

⇒ −x = 1 + 2

⇒ x = − 3

∴ x = −1 and −3

(iii)

If x < 1 then −(x − 1) = −(x − 3)

⇒ x − 1 = x − 3

This is an invalid equation.

If 1 < x < 3 then (x − 1) = −(x − 3)

⇒ x − 1 = −x + 3

⇒ x + x = 3 + 1

⇒ 2x = 4

⇒ x = 2

If x > 3 then (x − 1) = (x − 3)

This is again an invalid equation.

Thus, x = 2 satisfies the given equation.

(iv)

If x < 3 then −(x − 3) = −(x − 4)

⇒ x − 3 = x − 4

This is an invalid equation.

If 3 < x < 4 then (x − 3) = −(x − 4)

⇒ x − 3 = −x + 4

⇒ x + x = 4 + 3

⇒ 2x = 7

⇒ x =

If x > 4 then (x − 3) = (x − 4)

This is again an invalid equation.

Thus, x = satisfies the given equation.

(v)

If x < −2 then −(x + 2) = −(x − 5)

⇒ x + 2 = x − 5

This is an invalid equation.

If −2 < x < 5 then (x + 2) = −(x − 5)

⇒ x + 2 = −x + 5

⇒ x + x = 5 − 2

⇒ 2x = 3

⇒ x =

If x > 5 then (x + 2) = (x − 5)

This is again an invalid equation.

Thus, x = satisfies the given equation.

(vi)

If x < −1 then −x = −(x + 1)

⇒ x = x + 1

This is an invalid equation.

If −1 < x < 0 then −x = (x + 1)

⇒ −x = x + 1

⇒ x + x = −1

⇒ 2x = −1

⇒ x =

If x > 0 then x = (x + 1)

This is again an invalid equation.

Thus, x = satisfies the given equation.