Verify by substitution that
(i) the root of 3x − 5 = 7 is x = 4
(ii) the root of 3 + 2x = 9 is x = 3
(iii) the root of 5x − 8 = 2x − 2 is x = 2
(iv) the root of 8 − 7y = 1 is y = 1
(v) the root of $\frac{z}{7} = 8$ is z = 56

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Solution

(i)

$3 x - 5 = 7 S u b s t i t u t i n g x = 4 i n t h e g i v e n e q u a t i o n : L . H . S . : 3 \times 4 - 5 o r, 12 - 5 = 7 = R . H . S . L . H . S . = R . H . S . H e n c e, x = 4 i s t h e r o o t o f t h e g i v e n e q u a t i o n .$

(ii)

$3 + 2 x = 9 S u b s t i t u t i n g x = 3 i n t h e g i v e n e q u a t i o n : L . H . S . : 3 + 2 \times 3 o r, 3 + 6 = 9 = R . H . S . L . H . S . = R . H . S . H e n c e, x = 3 i s t h e r o o t o f t h e g i v e n e q u a t i o n .$

(iii)

$5 x - 8 = 2 x - 2 S u b s t i t u t i n g x = 2 i n t h e g i v e n e q u a t i o n : L . H . S . : R . H . S . : 5 \times 2 - 8 = 2 \times 2 - 2 o r, 10 - 8 = 2 = 4 - 2 = 2 L . H . S . = R . H . S . H e n c e, x = 2 i s t h e r o o t o f t h e g i v e n e q u a t i o n .$

(iv)

$8 - 7 y = 1 S u b s t i t u t i n g y = 1 i n t h e g i v e n e q u a t i o n : L . H . S . : 8 - 7 \times 1 o r, 8 - 7 = 1 = R . H . S . L . H . S . = R . H . S . H e n c e, y = 1 i s t h e r o o t o f t h e g i v e n e q u a t i o n .$

(v)

$\frac{z}{7} = 8 S u b s t i t u t i n g z = 56 i n t h e g i v e n e q u a t i o n : L . H . S . : \frac{56}{7} = 8 = R . H . S . L . H . S . = R . H . S . H e n c e, z = 56 i s t h e r o o t o f t h e g i v e n e q u a t i o n .$

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Similar questions

5 x - 8 = 2 x - 2

x = 2

\frac{3 x}{2} = 6

\frac{1}{2} x + 7 = 11

3 + 2 x = 9

x = 3

x^{3} - 5 x^{2} + 2 x + 8 = 0

\frac{z}{7} = 8

z = 56

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