Question

How do you solve the simultaneous equations $3x + 5y = 11$ and $2x + y = 3$?

Answer 1

Hint: In this question, a linear equation of two variables is given. Here we will use the substitution method to solve these two linear equations. To solve the equations using the substitution method, we should follow the below steps:
Select one equation and solve it for one of its variables.
On the other equation, substitute for the variable that we get from the first step.
Solve the new equation.
Substitute the value that we found into any equation involving both variables and solve for the other variable.
Check the solution in both original equations.

Complete step-by-step solution:
Here, we want to solve the equations using the substitution method.
$ \Rightarrow 3x + 5y = 11$ ……………....(1)
$ \Rightarrow 2x + y = 3$ ………………...(2)
The first step is to select one equation and solve it for one of its variables.
Let us take equation (1), and convert it into the form of x variable.
So, subtract 5y on both sides.
 $ \Rightarrow 3x + 5y - 5y = 11 - 5y$
That is equal to,
$ \Rightarrow 3x = 11 - 5y$
Now, divide both sides by 3.
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{11 - 5y}}{3}$
So, the value of x will be,
$ \Rightarrow x = \dfrac{{11 - 5y}}{3}$
In the second step, on the other equation, substitute for the variable that we get from the first step.
 Substitute $x = \dfrac{{11 - 5y}}{3}$ in equation (2).
 $ \Rightarrow 2x + y = 3$
$ \Rightarrow 2\left( {\dfrac{{11 - 5y}}{3}} \right) + y = 3$
That is equal to,
$ \Rightarrow \left( {\dfrac{{22 - 10y}}{3}} \right) + y = 3$
Let us take LCM to the left-hand side.
$ \Rightarrow \dfrac{{22 - 10y + 3y}}{3} = 3$
So,
$ \Rightarrow \dfrac{{22 - 7y}}{3} = 3$
Multiply by 3 on both sides.
$ \Rightarrow \left( {\dfrac{{22 - 7y}}{3}} \right) \times 3 = 3 \times 3$
So,
$ \Rightarrow 22 - 7y = 9$
Let us subtract -22 on both sides.
$ \Rightarrow 22 - 22 - 7y = 9 - 22$
$ \Rightarrow - 7y = - 13$
Now, divide by -7 into both sides.
$ \Rightarrow \dfrac{{ - 7y}}{{ - 7}} = \dfrac{{ - 13}}{{ - 7}}$
So,
$ \Rightarrow y = \dfrac{{13}}{7}$
Now, put the value of y in equation (1).
$ \Rightarrow 3x + 5y = 11$
Put the value of y.
$ \Rightarrow 3x + 5\left( {\dfrac{{13}}{7}} \right) = 11$
That is equal to
$ \Rightarrow 3x + \dfrac{{65}}{7} = 11$
Let us subtract $\dfrac{{65}}{7}$ to both sides.
 $ \Rightarrow 3x + \dfrac{{65}}{7} - \dfrac{{65}}{7} = 11 - \dfrac{{65}}{7}$
Now, take LCM to the right-hand side.
$ \Rightarrow 3x = \dfrac{{77 - 65}}{7}$
$ \Rightarrow 3x = \dfrac{{12}}{7}$
Now, divide both sides by 3.
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{12}}{{7 \times 3}}$
So,
$ \Rightarrow x = \dfrac{4}{7}$

Hence, we find the value of x as $\dfrac{4}{7}$ and the value of y as $\dfrac{{13}}{7}$ using substitution methods.

Note: To check whether our answer is correct or not, feed the x and y values in each equation.
Let us take equation (1) and put the values.
$ \Rightarrow 3x + 5y = 11$
Put $x = \dfrac{4}{7}$ and $y = \dfrac{{13}}{7}$.
$ \Rightarrow 3\left( {\dfrac{4}{7}} \right) + 5\left( {\dfrac{{13}}{7}} \right) = 11$
$ \Rightarrow \dfrac{{12}}{7} + \dfrac{{65}}{7} = 11$
Let us take LCM to the left-hand side.
$ \Rightarrow \dfrac{{12 + 65}}{7} = 11$
$ \Rightarrow \dfrac{{77}}{7} = 11$
That is equal to,
$ \Rightarrow 11 = 11$
Now, let us take equation (2) and put the values.
$ \Rightarrow 2x + y = 3$
Put $x = \dfrac{4}{7}$ and $y = \dfrac{{13}}{7}$.
$ \Rightarrow 2\left( {\dfrac{4}{7}} \right) + \left( {\dfrac{{13}}{7}} \right) = 3$
$ \Rightarrow \dfrac{8}{7} + \dfrac{{13}}{7} = 3$
Let us take LCM to the left-hand side.
$ \Rightarrow \dfrac{{8 + 13}}{7} = 3$
$ \Rightarrow \dfrac{{21}}{7} = 3$
That is equal to,
$ \Rightarrow 3 = 3$