Question
Solve the following equation and verify your answer:
$\Big(\frac{\text{x}+1}{\text{x}-4}\Big)^2=\frac{\text{x}+8}{\text{x}-2}$
$\Big(\frac{\text{x}+1}{\text{x}-4}\Big)^2=\frac{\text{x}+8}{\text{x}-2}$
✓
Answer
$\Big(\frac{\text{x}+1}{\text{x}-4}\Big)^2=\frac{\text{x}+8}{\text{x}-2}$
$\Rightarrow\frac{(\text{x}+1)^2}{(\text{x}-4)^2}=\frac{(\text{x}+8)}{(\text{x}-2)}$
By cross multiplication,
$=(\text{x}+1)^2(\text{x}-2)=(\text{x}-4)^2(\text{x}+8)$
$\Rightarrow(\text{x}^2+2\text{x}+1)(\text{x}-2)=(\text{x}^2-8\text{x}+16)(\text{x}+8)$
$\Rightarrow\text{x}^3-2\text{x}^2+2\text{x}^2-4\text{x}+\text{x}-2\\=\text{x}^3+8\text{x}^2-8\text{x}^2-64\text{x}+16\text{x}+128$
$\Rightarrow\text{x}^3-3\text{x}-2=\text{x}^3-48\text{x}+128$
$\Rightarrow\text{x}^3 -3\text{x}-\text{x}^3+48\text{x}-128+2$
(By transposition)
$\Rightarrow45\text{x}$
$=130$
$\Rightarrow\text{x}=\frac{130}{45}$
$=\frac{26}{9}$
$\therefore\text{x}=\frac{26}{9}$
Verification:
$\text{L.H.S}=\bigg(\frac{\text{x}+1}{\text{x}-4}\bigg)^2=\Bigg(\frac{\frac{26}{9}+1}{\frac{26}{9}-4}\Bigg)^2$
$=\Bigg(\frac{\frac{26+9}{9}}{\frac{26-36}{9}}\Bigg)^2=\Bigg(\frac{\frac{35}{9}}{\frac{-10}{9}}\Bigg)^2$
$=\bigg(\frac{35}{9}\times\frac{9}{-10}\bigg)^2=\bigg(\frac{-7}{2}\bigg)^2=\frac{49}{4}$
$\text{R.H.S}=\frac{\text{x}+8}{\text{x}-2}=\frac{\frac{26}{9}+8}{\frac{26}{9}-2}=\frac{\frac{26+72}{9}}{\frac{26-18}{9}}=\frac{\frac{98}{9}}{\frac{8}{9}}$
$=\frac{98}{9}\times\frac{9}{8}=\frac{49}{4}$
$\therefore\text{L.H.S}=\text{R.H.S}$
$\Rightarrow\frac{(\text{x}+1)^2}{(\text{x}-4)^2}=\frac{(\text{x}+8)}{(\text{x}-2)}$
By cross multiplication,
$=(\text{x}+1)^2(\text{x}-2)=(\text{x}-4)^2(\text{x}+8)$
$\Rightarrow(\text{x}^2+2\text{x}+1)(\text{x}-2)=(\text{x}^2-8\text{x}+16)(\text{x}+8)$
$\Rightarrow\text{x}^3-2\text{x}^2+2\text{x}^2-4\text{x}+\text{x}-2\\=\text{x}^3+8\text{x}^2-8\text{x}^2-64\text{x}+16\text{x}+128$
$\Rightarrow\text{x}^3-3\text{x}-2=\text{x}^3-48\text{x}+128$
$\Rightarrow\text{x}^3 -3\text{x}-\text{x}^3+48\text{x}-128+2$
(By transposition)
$\Rightarrow45\text{x}$
$=130$
$\Rightarrow\text{x}=\frac{130}{45}$
$=\frac{26}{9}$
$\therefore\text{x}=\frac{26}{9}$
Verification:
$\text{L.H.S}=\bigg(\frac{\text{x}+1}{\text{x}-4}\bigg)^2=\Bigg(\frac{\frac{26}{9}+1}{\frac{26}{9}-4}\Bigg)^2$
$=\Bigg(\frac{\frac{26+9}{9}}{\frac{26-36}{9}}\Bigg)^2=\Bigg(\frac{\frac{35}{9}}{\frac{-10}{9}}\Bigg)^2$
$=\bigg(\frac{35}{9}\times\frac{9}{-10}\bigg)^2=\bigg(\frac{-7}{2}\bigg)^2=\frac{49}{4}$
$\text{R.H.S}=\frac{\text{x}+8}{\text{x}-2}=\frac{\frac{26}{9}+8}{\frac{26}{9}-2}=\frac{\frac{26+72}{9}}{\frac{26-18}{9}}=\frac{\frac{98}{9}}{\frac{8}{9}}$
$=\frac{98}{9}\times\frac{9}{8}=\frac{49}{4}$
$\therefore\text{L.H.S}=\text{R.H.S}$
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