Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Solve the following pairs of linear equations by elimination method: $47x + 31y = 63$ and $31x + 47y = 15$A. $x = 1,y = 0$B. $x = 2,y = - 1$C. $x = 3,y = 4$D. $x = 9,y = - 7$

Last updated date: 18th Sep 2024
Total views: 413.1k
Views today: 5.13k
Verified
413.1k+ views
Hint: Here, we are required to solve the given pairs of linear equations by elimination method. We will try making the coefficients of either of the two variables same in both the equations and then, we will add or subtract one equation from the other to find the value of one variable. We will substitute the obtained variable in one of the given equations to find the value of the other variable, hence, giving us the solution of the given pairs of linear equations.

According to the question, we are given two equations:
$47x + 31y = 63$………………………. $\left( 1 \right)$
$31x + 47y = 15$….…………………… $\left( 2 \right)$
Now, since, we are required to solve these equations using elimination methods.
Hence, multiplying equation $\left( 1 \right)$ by 31 and equation $\left( 2 \right)$ by 47, we get,
Equation 1:
$31\left( {47x + 31y} \right) = 31 \times 63$
$\Rightarrow 1457x + 961y = 1953$………………….. $\left( 3 \right)$
Equation 2:
$47\left( {31x + 47y} \right) = 47 \times 15$
$\Rightarrow 1457x + 2209y = 705$………………….. $\left( 4 \right)$
Now, subtracting the equation $\left( 4 \right)$ from the equation $\left( 3 \right)$ we get
$\Rightarrow 1457x + 961y - 1457x - 2209y = 1953 - 705$
$\Rightarrow - 1248y = 1248$
Dividing both sides by $- 1248$, we get
$\Rightarrow y = - 1$
Now, substituting $y = - 1$ in equation $\left( 1 \right)$, we get
$47x + 31\left( { - 1} \right) = 63$
$\Rightarrow 47x - 31 = 63$
Adding 31 on both sides, we get
$\Rightarrow 47x = 63 + 31 = 94$
Dividing both sides by 47, we get
$\Rightarrow x = 2$
Therefore, the required solution of the given equations $47x + 31y = 63$ and $31x + 47y = 15$ is $x = 2$ and $y = - 1$.
Hence, option B is the correct answer.

Note: An alternate way to solve these equations using the elimination method only is:
First of all, we should add both the equations $\left( 1 \right)$ and $\left( 2 \right)$,
$\Rightarrow 47x + 31y + 31x + 47y = 63 + 15$
$\Rightarrow 78x + 78y = 78$
Dividing both sides by 78,
$\Rightarrow x + y = 1$……………………………. $\left( 5 \right)$
Now, similarly, subtracting the equation $\left( 2 \right)$from $\left( 1 \right)$, we get,
$\Rightarrow 47x + 31y - 31x - 47y = 63 - 15$
$\Rightarrow 16x - 16y = 48$
Dividing both sides by 16, we get
$\Rightarrow x - y = 3$…………………………. $\left( 6 \right)$
Now, by the elimination method, adding the equations $\left( 5 \right)$and $\left( 6 \right)$, we get,
$\Rightarrow x + y + x - y = 1 + 3$
$\Rightarrow 2x = 4$
Dividing both sides by 2, we get
$\Rightarrow x = 2$
Now, substituting the value of $x = 2$in $\left( 5 \right)$, we get,
$\Rightarrow 2 + y = 1$
Subtracting 2 from both sides, we get
$\Rightarrow y = 1 - 2 = - 1$
Therefore, the required solution of the given equations $47x + 31y = 63$ and $31x + 47y = 15$is $x = 2$ and $y = - 1$.
Hence, option B is the correct answer.