EXERCISE 8.1

1. Verify by substitution that:

(i) [tex]\( x = 4 \)[/tex] is the root of [tex]\( 3x - 5 = 7 \)[/tex]

(ii) [tex]\( x = 3 \)[/tex] is the root of [tex]\( 5 + 3x = 14 \)[/tex]

(iii) [tex]\( x = 2 \)[/tex] is the root of [tex]\( 3x - 2 = 8x - 12 \)[/tex]

(iv) [tex]\( x = 4 \)[/tex] is the root of [tex]\( \frac{3x}{2} = 6 \)[/tex]

(v) [tex]\( y = 2 \)[/tex] is the root of [tex]\( y - 3 = 2y - 5 \)[/tex]

(vi) [tex]\( x = 8 \)[/tex] is the root of [tex]\( \frac{1}{2}x + 7 = 11 \)[/tex]



Answer :

Let's verify the given roots by substituting them into their respective equations.

### Part (i)
We need to verify if [tex]\( x = 4 \)[/tex] is the root of [tex]\( 3x - 5 = 7 \)[/tex].

Substitute [tex]\( x = 4 \)[/tex]:
[tex]\[ 3(4) - 5 = 12 - 5 = 7 \][/tex]

Left-hand side (LHS) = Right-hand side (RHS)

Since both sides are equal, [tex]\( x = 4 \)[/tex] is a root of the equation [tex]\( 3x - 5 = 7 \)[/tex].

### Part (ii)
We need to verify if [tex]\( x = 3 \)[/tex] is the root of [tex]\( 5 + 3x = 14 \)[/tex].

Substitute [tex]\( x = 3 \)[/tex]:
[tex]\[ 5 + 3(3) = 5 + 9 = 14 \][/tex]

LHS = RHS

Since both sides are equal, [tex]\( x = 3 \)[/tex] is a root of the equation [tex]\( 5 + 3x = 14 \)[/tex].

### Part (iii)
We need to verify if [tex]\( x = 2 \)[/tex] is the root of [tex]\( 3x - 2 = 8x - 12 \)[/tex].

Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[ 3(2) - 2 = 6 - 2 = 4 \][/tex]
[tex]\[ 8(2) - 12 = 16 - 12 = 4 \][/tex]

LHS = RHS

Since both sides are equal, [tex]\( x = 2 \)[/tex] is a root of the equation [tex]\( 3x - 2 = 8x - 12 \)[/tex].

### Part (iv)
We need to verify if [tex]\( x = 4 \)[/tex] is the root of [tex]\( \frac{3x}{2} = 6 \)[/tex].

Substitute [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{3(4)}{2} = \frac{12}{2} = 6 \][/tex]

LHS = RHS

Since both sides are equal, [tex]\( x = 4 \)[/tex] is a root of the equation [tex]\( \frac{3x}{2} = 6 \)[/tex].

### Part (v)
We need to verify if [tex]\( y = 2 \)[/tex] is the root of [tex]\( y - 3 = 2y - 5 \)[/tex].

Substitute [tex]\( y = 2 \)[/tex]:
[tex]\[ 2 - 3 = -1 \][/tex]
[tex]\[ 2(2) - 5 = 4 - 5 = -1 \][/tex]

LHS = RHS

Since both sides are equal, [tex]\( y = 2 \)[/tex] is a root of the equation [tex]\( y - 3 = 2y - 5 \)[/tex].

### Part (vi)
We need to verify if [tex]\( x = 8 \)[/tex] is the root of [tex]\( \frac{1}{2}x + 7 = 11 \)[/tex].

Substitute [tex]\( x = 8 \)[/tex]:
[tex]\[ \frac{1}{2}(8) + 7 = 4 + 7 = 11 \][/tex]

LHS = RHS

Since both sides are equal, [tex]\( x = 8 \)[/tex] is a root of the equation [tex]\( \frac{1}{2}x + 7 = 11 \)[/tex].

All given roots are verified to satisfy their respective equations.