Question

Verify by substitution that:
(i) x = 4 is the root of 3x − 5 = 7
(ii) x = 3 is the root of 5 + 3x = 14
(iii) x = 2 is the root of 3x − 2 = 8x − 12
(iv) x = 4 is the root of $\frac{3x}{2}=6$
(v) y = 2 is the root of y − 3 = 2y − 5
(vi) x = 8 is the root of $\frac{1}{2}x+7=11$

Answer 1

(i) x = 4 is the root of 3x − 5 = 7.
Now, substituting x = 4 in place of 'x' in the given equation 3x − 5 = 7,
3(4) − 5 = 7
12 − 5 = 7
7 = 7
LHS = RHS
Hence, x = 4 is the root of 3x − 5 = 7.

(ii) x = 3 is the root of 5 + 3x = 14.
Now, substituting x = 3 in place of 'x' in the given equation 5 + 3x = 14,
5 + 3(3) = 14
5 + 9 = 14
14 = 14
LHS = RHS
Hence, x = 3 is the root of 5 + 3x = 14.

(iii) x = 2 is the root of 3x − 2 = 8x − 12.
Now, substituting x = 2 in place of 'x' in the given equation 3x − 2 = 8x − 12,
3(2) − 2 = 8(2) − 12
6 − 2 = 16 − 12
4 = 4
LHS = RHS
Hence, x = 2 is the root of 3x − 2 = 8x − 12.

(iv) x = 4 is the root of $\frac{3x}{2}=6$.
Now, substituting x = 4 in place of 'x' in the given equation $\frac{3x}{2}=6$,
$$\begin{gathered} \frac{3\times 4}{2}=6 \\ \frac{12}{2}=6 \\ 6=6 \end{gathered}$$
LHS = RHS
Hence, x = 4 is the root of $\frac{3x}{2}=6$.
(v) y = 2 is the root of y − 3 = 2y − 5.
Now, substituting y = 2 in place of 'y' in the given equation y − 3 = 2y − 5,
2 − 3 = 2(2) − 5
−1 = 4 − 5
−1 = −1
LHS = RHS
Hence, y = 2 is the root of y − 3 = 2y − 5.

(vi) x = 8 is the root of $\frac{1}{2}x+7=11$.
Now, substituting x = 8 in place of 'x' in the given equation $\frac{1}{2}x+7=11$,
$\frac{1}{2}\times 8+7=11$
4 + 7 = 11
11 = 11
LHS = RHS
Hence, x = 8 is the root of $\frac{1}{2}x+7=11$.