Question

# Solve the equation $7x+12y=220$ , in positive integers.(a).$\left( 2,24 \right)$ (b).$\left( 28,2 \right)$ (c).$(32,3)$ (d).$\left( 2,34 \right)$

Hint: Here we have an equation with two variables. To find out the solution try to establish a relation between the variables and check the options one by one.

Let us first take the given equation:
$7x+12y=220...........(1)$
Now look at the equation very carefully, there are two variables. Variable is basically a symbol for a number we don’t know yet or we can say unknown. Here $x,y$ are unknowns to us. We have to find out some specific integer values for $x,y$.
Generally for two variables if we have two equations we get a unique solution. Here we have only one equation but two variables. So basically if we put any integer value for one variable then we will get a value for another variable.
Now, let us find out the relation between $x$ and $y$ .
The equation is:
$7x+12y=220$
Take $12y$ from left side to right side:
$\Rightarrow 7x=220-12y$
Divide both the sides by 7:
$\Rightarrow \dfrac{7x}{7}=\dfrac{220-12y}{7}$
$\Rightarrow x=\dfrac{220-12y}{7}......(2)$
If we put any value for $y$ we will always get a value of $x$ .
Here we have four options. So, we will put the values of y from the options one by one and we will check if the value of $x$ is correct or not.
Our first option is $\left( 2,24 \right)$ . So here $y=24$
Let us put the value of y in equation (2)
\begin{align} & x=\dfrac{220-\left( 12\times 24 \right)}{7} \\ & \Rightarrow x=\dfrac{220-288}{7} \\ & \Rightarrow x=\dfrac{-8}{7} \\ \end{align}
So for $y=24$ , $x\ne 2$ . Hence option (a) is not correct.
Our second option is $\left( 28,2 \right)$ . So here $y=2$
Let us put the value of y in equation (2)
\begin{align} & x=\dfrac{220-\left( 12\times 2 \right)}{7} \\ & \Rightarrow x=\dfrac{220-24}{7} \\ & \Rightarrow x=\dfrac{196}{7}=28 \\ \end{align}
So for $y=2$ , $x=28$ . Hence option (b) is correct.
Our third option is $\left( 32,3 \right)$ . So here $y=3$
Let us put the value of y in equation (2)
\begin{align} & x=\dfrac{220-\left( 12\times 3 \right)}{7} \\ & \Rightarrow x=\dfrac{220-36}{7} \\ & \Rightarrow x=\dfrac{184}{7}=26\dfrac{2}{7} \\ \end{align}
So for $y=3$ , $x\ne 32$ . Hence option (c) is not correct.
Our fourth option is $\left( 2,34 \right)$ . So here $y=34$
Let us put the value of y in equation (2)
\begin{align} & x=\dfrac{220-\left( 12\times 34 \right)}{7} \\ & \Rightarrow x=\dfrac{220-408}{7} \\ & \Rightarrow x=\dfrac{-188}{7} \\ \end{align}
So for $y=34$ , $x\ne 2$ . Hence option (d) is not correct.
Therefore, option (b) is the correct answer.

Note: We can also directly put the values from the options in the left hand side of the equation:
$7x+12y=220$ , and check if it is coming 220 or not.
Like if we substitute $\left( 28,2 \right)$ in the left hand side we will get:
$=\left( 7\times 28 \right)+\left( 12\times 2 \right)=196+24=220$
Hence, option (b) is correct.