Question

Put the correct number in the box, so that the equation remains true.

If $\text{11y-100=y+10}$ then $11y-100=\frac{y+10}{\boxed{}}$

Answer 1

Step 1: Solving given condition for $y$

Given, If $\text{11y-100=y+10}$

$\text{⇒11y-100=y+10}$

$$\begin{gathered} \Rightarrow 11\mathrm{y}-100+100=\mathrm{y}+10+100 \\ \Rightarrow 11\mathrm{y}=\mathrm{y}+110 \end{gathered}$$

$\Rightarrow 11\mathrm{y}-\mathrm{y}=\mathrm{y}+110-\mathrm{y}$

$\Rightarrow 10\mathrm{y}=110$

$\Rightarrow \mathrm{y}=11$

Step 2: Solving the given question using $\mathrm{y}=11$

Let we have to put $\mathrm{x}$ in the box so that the equation remains true

$\Rightarrow 11\mathrm{y}-100=\frac{\mathrm{y}+10}{\mathrm{x}}$

Put $\mathrm{y}=11$ in both sides of the given expression:

$$\begin{gathered} \Rightarrow 11\times 11-100=\frac{11+10}{\mathrm{x}} \\ \Rightarrow 121-100=\frac{21}{\mathrm{x}} \\ \Rightarrow 21=\frac{21}{\mathrm{x}} \end{gathered}$$

$$\begin{gathered} \Rightarrow 21\mathrm{x}=\frac{21}{\bcancel{\mathrm{x}}}\times \bcancel{\mathrm{x}}[\mathrm{Multiply}\mathrm{both}\mathrm{side}\mathrm{by}\mathrm{x}] \\ \Rightarrow \frac{\bcancel{21}\mathrm{x}}{\bcancel{21}}=\frac{\bcancel{21}}{\bcancel{21}}[\mathrm{Dividing}\mathrm{both}\mathrm{side}\mathrm{by}21] \\ \therefore \mathrm{x}=1 \end{gathered}$$

Therefore we place $1$ in the box

Hence, If $\text{11y-100=y+10}$ then $11y-100=\frac{y+10}{\boxed{1}}$.