Question
Solve the following equations :
$8 x+4=3(x-1)+7.$
$8 x+4=3(x-1)+7.$
✓
Answer
We have,
$8 x+4=3(x-1)+7$
$\Rightarrow \quad 8 x+4=3 x-3+7$
$\Rightarrow \quad 8 x+4=3 x+4$
$\Rightarrow \quad 8 x-3 x=4-4\quad$ [transposing $3 x$ to LHS and 4 to RHS]
$\Rightarrow \quad 5 x=0$
$\Rightarrow \quad x=\frac{0}{5} \quad$ [dividing both sides by 5]
$\therefore x=0$, which is the required solution.
Check For x = 0,
$L H S=8(0)+4=4$
RHS $=3(0-1)+7=4$
$\therefore \quad$ LHS $=$ RHS $\quad$ [as required]
$8 x+4=3(x-1)+7$
$\Rightarrow \quad 8 x+4=3 x-3+7$
$\Rightarrow \quad 8 x+4=3 x+4$
$\Rightarrow \quad 8 x-3 x=4-4\quad$ [transposing $3 x$ to LHS and 4 to RHS]
$\Rightarrow \quad 5 x=0$
$\Rightarrow \quad x=\frac{0}{5} \quad$ [dividing both sides by 5]
$\therefore x=0$, which is the required solution.
Check For x = 0,
$L H S=8(0)+4=4$
RHS $=3(0-1)+7=4$
$\therefore \quad$ LHS $=$ RHS $\quad$ [as required]
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