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# The pair of linear equations $x + 2y = 5$, $7x + 3y = 13$ has a unique solutionA. $x = 1$, $y = 2$B. $x = 2$, $y = 1$C. $x = 3$, $y = 1$D. $x = 1$, $y = 3$

Last updated date: 19th Apr 2024
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Hint: Here, we will use the method of elimination for solving a system of two equations by multiplication and addition to eliminate a variable from one equation. Then we can find the value of the remaining variable by substitution in the original equation.

Complete step-by-step solution:
Given that the equations are
$x + 2y = 5{\text{ ......}}\left( 1 \right)$
$7x + 3y = 13{\text{ ......}}\left( 2 \right)$

Multiplying the equation $\left( 1 \right)$ by 7 and equation $\left( 2 \right)$ by $- 1$, we get
$\left( {x + 2y = 5} \right) \times 7 \Rightarrow 7x + 14y = 35{\text{ ......}}\left( 3 \right)$
$\left( {7x + 3y = 13} \right) \times - 1 \Rightarrow - 7x - 3y = - 13{\text{ ......}}\left( 4 \right)$

Adding equation $\left( 3 \right)$ and equation $\left( 4 \right)$, we get

$\underline {7x + 14y = 35 \\ - 7x - 3y = - 13 \\ } \\ {\text{ }}11y = 22 \\$

Dividing this equation by 2, we get

$\Rightarrow y = 2$

Substituting this value of $y$ in the equation $\left( 1 \right)$, we get

$\Rightarrow x + 2\left( 2 \right) = 5 \\ \Rightarrow x = 5 - 4 \\ \Rightarrow x = 1 \\$

Thus, we have $x = 1$ and $y = 2$.

Hence, the option A is correct.

Note: 1.In the system of three equations we can also solve this using substitution method if the method to be followed is not mentioned in the question.
2.We will substitute the values of $x$ and $y$ in the original three equations to verify the solution.
$1 + 2\left( 2 \right) = 5 \\ 1 + 4 = 5 \\ 5 = 5 \\$

$7\left( 1 \right) + 3\left( 2 \right) = 13 \\ 7 + 6 = 13 \\ 13 = 13 \\$

Since ${\text{LHS}} = {\text{RHS}}$ in all the above equations, we have verified that the solution is correct.