Question

How do you solve this system of equations:
$2x + 3y = 8$ and $3x + 1y = 5$ ?

Answer 1

Hint: To solve such questions start by multiplying the second given equation with the number $3$ . Then subtract the first given equation from the other one which is multiplied by the number $3$ . Once the value of $x$ is found use that to find the value of $y$ .

Complete step by step answer:
Given the equations $2x + 3y = 8$ and $3x + 1y = 5$ .
It is asked to solve these equations.
Suppose that
$2x + 3y = 8......(1)$
And
$3x + 1y = 5......(2)$
Multiply both the LHS and RHS of the equation $\left( 2 \right)$ with the number $3$ , that is
$\left( {3x + 1y} \right) \times 3 = 5 \times 3$
Further computing we get,
$9x + 3y = 15......(3)$
Next, subtract the equation $\left( 1 \right)$ from the equation $\left( 3 \right)$ , that is
$2x + 3y = 8 -$
$9x + 3y = 15$
Further simplifying we get,
$- 7x = - 7$
Dividing throughout by number $- 7$ , we get
$\dfrac{{ - 7x}}{{ - 7}} = \dfrac{{ - 7}}{{ - 7}}$
That is we get
$x = 1$
Now substitute the value of $x$ in equation $\left( 2 \right)$ , that is
$3\left( 1 \right) + 1y = 5$
Further simplifying, we get
$3 + y = 5$
That is,
$y = 5 - 3$
Subtracting the RHS we get,
$y = 2$

Therefore, the solution of the given system of equations is $x = 1$ and $y = 2$ .

Additional information:
An equation that can be written in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are real numbers, and $a$ and $b$ are not both zero, is known as a linear equation in two variables $x$ and $y$ . Every solution of the equation is a point on the line representing it.

Note: These types of questions are easy to solve. There are different methods to solve linear equations with two variables, that is, method of substitution and method of elimination. In this solution part, we have used a method of elimination. One can check whether the obtained values of $x$ and $y$ is correct or not by putting those in the given equations.