Question
Solve the following equations. Check your result in case. $\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}=\frac{4\text{x}+1}{7}$
✓
Answer
$\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}=\frac{4\text{x}+1}{7}$
$\frac{28(2\text{x}-3)+35(\text{x}+3)=20(4\text{x}+1)}{140}$
$(LCM$ of $5, 4, 7 = 140)$ $28(2\text{x}-3)+35(\text{x}+3)=20(4\text{x}+1)$
$\Rightarrow56\text{x}-84+35\text{x}+105=80\text{x}+20$
$\Rightarrow56\text{x}+35\text{x}-80\text{x}=20+84-105$
$\Rightarrow91\text{x}-80\text{x}=140-105$
$\Rightarrow11\text{x}=-1$
$\Rightarrow\text{x}=\frac{-1}{11}$
$\therefore\text{x}=\frac{-1}{11}$
Check: $\text{L.H.S.}=\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}$
$=\frac{2\Big(-\frac{1}{11}\Big)-3}{5}+\frac{\frac{-1}{11}+3}{4}$
$=\frac{\frac{-2}{11}-3}{5}+\frac{\frac{-1}{11}+3}{4}$
$=\frac{-2-33}{11\times5}+\frac{-1+33}{11\times4}$
$=\frac{-35}{11\times5}+\frac{32}{11\times4}$
$=\frac{-7}{11}+\frac{8}{11}$
$=\frac{1}{11}$
$\text{R.H.S}.=\frac{4\text{x}+1}{7}$
$=\frac{4\Big(-\frac{1}{11}\Big)+1}{7}$
$=\frac{-\frac{4}{11}+1}{7}$
$=\frac{-4+11}{11\times7}$
$=\frac{7}{11\times7}$
$=\frac{1}{11}$
$\text{L.H.S. = R.H.S.}$
$\frac{28(2\text{x}-3)+35(\text{x}+3)=20(4\text{x}+1)}{140}$
$(LCM$ of $5, 4, 7 = 140)$ $28(2\text{x}-3)+35(\text{x}+3)=20(4\text{x}+1)$
$\Rightarrow56\text{x}-84+35\text{x}+105=80\text{x}+20$
$\Rightarrow56\text{x}+35\text{x}-80\text{x}=20+84-105$
$\Rightarrow91\text{x}-80\text{x}=140-105$
$\Rightarrow11\text{x}=-1$
$\Rightarrow\text{x}=\frac{-1}{11}$
$\therefore\text{x}=\frac{-1}{11}$
Check: $\text{L.H.S.}=\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}$
$=\frac{2\Big(-\frac{1}{11}\Big)-3}{5}+\frac{\frac{-1}{11}+3}{4}$
$=\frac{\frac{-2}{11}-3}{5}+\frac{\frac{-1}{11}+3}{4}$
$=\frac{-2-33}{11\times5}+\frac{-1+33}{11\times4}$
$=\frac{-35}{11\times5}+\frac{32}{11\times4}$
$=\frac{-7}{11}+\frac{8}{11}$
$=\frac{1}{11}$
$\text{R.H.S}.=\frac{4\text{x}+1}{7}$
$=\frac{4\Big(-\frac{1}{11}\Big)+1}{7}$
$=\frac{-\frac{4}{11}+1}{7}$
$=\frac{-4+11}{11\times7}$
$=\frac{7}{11\times7}$
$=\frac{1}{11}$
$\text{L.H.S. = R.H.S.}$
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