Question
$8(2x - 5) - 6(3x - 7) = 1$
✓
Answer
$8(2x - 5) - 6(3x - 7) = 1$ On expanding the brackets, we get $(8 × 2x) - (8 × 5) - (6 × 3x) + (-6) × (-7) = 1$
$16x - 40 - 18x + 42 = 1$
$16x - 18x + 42 - 40 = 1$
$-2x + 2 = 1$
Subtracting $2$ from both sides, we get
$-2x+ 2 - 2 = 1 - 2$
$-2x = -1$
Multiplying both sides by $-1$, we get
$-2x × (-1) = -1 × (-1)$
$2x = 1$
Dividing both sides by $2$, we get
$\frac{2\text{x}}{2}=\frac12$
$\text{x}=\frac12$
Verification:
Substituting $\text{x}=\frac12$ in $L.H.S.$, we get
$=8\Big(2\times\frac{1}{2}-5\Big)-6\Big(3\times\frac{1}{2}-7\Big)$
$= 8(1 - 5) - 6(32 - 7)$
$= 8× (-4) - (6 \times 32) + (6 \times 7) $
$= -32 - 9 + 42 $
$= - 41 + 42 = 1 = \text{R.H.S.}$
$\text{L.H.S.} = \text{R.H.S.}$
Hence, verified.
$16x - 40 - 18x + 42 = 1$
$16x - 18x + 42 - 40 = 1$
$-2x + 2 = 1$
Subtracting $2$ from both sides, we get
$-2x+ 2 - 2 = 1 - 2$
$-2x = -1$
Multiplying both sides by $-1$, we get
$-2x × (-1) = -1 × (-1)$
$2x = 1$
Dividing both sides by $2$, we get
$\frac{2\text{x}}{2}=\frac12$
$\text{x}=\frac12$
Verification:
Substituting $\text{x}=\frac12$ in $L.H.S.$, we get
$=8\Big(2\times\frac{1}{2}-5\Big)-6\Big(3\times\frac{1}{2}-7\Big)$
$= 8(1 - 5) - 6(32 - 7)$
$= 8× (-4) - (6 \times 32) + (6 \times 7) $
$= -32 - 9 + 42 $
$= - 41 + 42 = 1 = \text{R.H.S.}$
$\text{L.H.S.} = \text{R.H.S.}$
Hence, verified.
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→| S.No. | Equation | Value | Say, whether the equation is satisfied (Yes/No) |
| (i) | x+3=0 | x=3 | |
| (ii) | x+3=0 | x=0 | |
| (iii) | x+3=0 | x=-3 | |
| (iv) | x-7=1 | x=7 | |
| (v) | x-7=1 | x=8 | |
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| (vii) | 5x=25 | x=5 | |
| (viii) | 5x=25 | x=-5 | |
| (ix) | $\frac{m}{2}$=2 | m=-6 | |
| (x) | $\frac{m}{2}$=2 | m=0 | |
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