Question 13 Marks
Write two equations for which $2$ is the solution.
Answer
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Let the two numbers be $x$ and $y,$ which has solution $2$ in equation.
$a.\ $For getting first equation, the number $x$ is multiplied by $2$, then the number is $2x$.
After that, $3$ is subtracted from it which results into $1$.
Hence, $2x - 3 = 1$
$2x = 3 + 1$ [transposing $-3$ to $RHS]$
$\Rightarrow 2x = 4$
$\Rightarrow\frac{2\text{x}}{2}=\frac{4}{2}$
$\Rightarrow x = 2$ [dividing both sides by $2]$
$b.\ $For getting second equation, the number $y$ is multiplied by $3$, then the number is $3y$. After that, it will be added to $4$ which results into $10.$
Hence, $3y + 4 = 10$
On solving, $3y = 10 - 4$
$\Rightarrow 3y = 6$ [transposing $+4$ to $RHS$]
$\Rightarrow\frac{3\text{y}}{3}=\frac{6}{3}$ [dividing both sides by $3]$
$\Rightarrow y = 2$
Hence, two equations are
$2x - 3 = 1$ and $3y + 4 = 10.$
$a.\ $For getting first equation, the number $x$ is multiplied by $2$, then the number is $2x$.
After that, $3$ is subtracted from it which results into $1$.
Hence, $2x - 3 = 1$
$2x = 3 + 1$ [transposing $-3$ to $RHS]$
$\Rightarrow 2x = 4$
$\Rightarrow\frac{2\text{x}}{2}=\frac{4}{2}$
$\Rightarrow x = 2$ [dividing both sides by $2]$
$b.\ $For getting second equation, the number $y$ is multiplied by $3$, then the number is $3y$. After that, it will be added to $4$ which results into $10.$
Hence, $3y + 4 = 10$
On solving, $3y = 10 - 4$
$\Rightarrow 3y = 6$ [transposing $+4$ to $RHS$]
$\Rightarrow\frac{3\text{y}}{3}=\frac{6}{3}$ [dividing both sides by $3]$
$\Rightarrow y = 2$
Hence, two equations are
$2x - 3 = 1$ and $3y + 4 = 10.$