Question
$\frac{\text{5x}-1}{3}-\frac{2\text{x}-2}{3}=1$
✓
Answer
$\frac{\text{5x}-1}{3}-\frac{2\text{x}-2}{3}=1$
$\frac{5\text{x}-1-2\text{x}+2}{3}=1$
$\frac{3\text{x}+1}{3}=1$
Multiplying both sides by 3, we get 3
$\frac{3\text{x}+1}{3}\times3=1\times3$
$=\text{3x}+1=3$
Subtracting 1 from both sides, we get
= 3x + 1 - 1 = 3 - 1
= 3x = 2
Dividing both sides by 3, we get
= 3x + 1 - 1 = 3 -1
= 3x = 2
Dividing both sides by 3, we get
$=\frac{\text{3x}}{3}=\frac{2}{3}$
$=\text{x}=\frac23$
Verification:
Substituting $\text{x}=\frac23$ in L.H.S., we get
$=\frac{5\big(\frac23\big)-1}{3}-\frac{2\big(\frac23\big)-2}{3}$
$=\frac{\frac{1}3-1}{3}-\frac{\frac43-3}{3}$
$=\frac{\frac{10-3}{3}}{3}-\frac{\frac{4-6}{3}}{3}$
$=\frac{7}{3\times3}-\Big(\frac{-2}{3\times3}\Big)$
$=\frac79+\frac29$
$=\frac99=1=\text{R.H.S.}$
L.H.S. = R.H.S.
Hence, verified.
$\frac{5\text{x}-1-2\text{x}+2}{3}=1$
$\frac{3\text{x}+1}{3}=1$
Multiplying both sides by 3, we get 3
$\frac{3\text{x}+1}{3}\times3=1\times3$
$=\text{3x}+1=3$
Subtracting 1 from both sides, we get
= 3x + 1 - 1 = 3 - 1
= 3x = 2
Dividing both sides by 3, we get
= 3x + 1 - 1 = 3 -1
= 3x = 2
Dividing both sides by 3, we get
$=\frac{\text{3x}}{3}=\frac{2}{3}$
$=\text{x}=\frac23$
Verification:
Substituting $\text{x}=\frac23$ in L.H.S., we get
$=\frac{5\big(\frac23\big)-1}{3}-\frac{2\big(\frac23\big)-2}{3}$
$=\frac{\frac{1}3-1}{3}-\frac{\frac43-3}{3}$
$=\frac{\frac{10-3}{3}}{3}-\frac{\frac{4-6}{3}}{3}$
$=\frac{7}{3\times3}-\Big(\frac{-2}{3\times3}\Big)$
$=\frac79+\frac29$
$=\frac99=1=\text{R.H.S.}$
L.H.S. = R.H.S.
Hence, verified.
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