Question

Solve the following linear system $3x - 2y = 4$, and $x + 4y = 3$?

Answer 1

Hint: 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 answer:
Here, we want to solve the equations using the substitution method.
$ \Rightarrow 3x - 2y = 4$ ...(1)
$ \Rightarrow x + 4y = 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, add 2y on both sides.
 $ \Rightarrow 3x - 2y + 2y = 4 + 2y$
That is equal to,
$ \Rightarrow 3x = 4 + 2y$
Now, divide both sides by 3.
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{4 + 2y}}{3}$
So, the value of x will be,
$ \Rightarrow x = \dfrac{{4 + 2y}}{3}$
In the second step, on the other equation, substitute for the variable that we get from the first step.
 Substitute $x = \dfrac{{4 + 2y}}{3}$ in equation (2).
 $ \Rightarrow x + 4y = 3$
$ \Rightarrow \left( {\dfrac{{4 + 2y}}{3}} \right) + 4y = 3$
Let us take LCM on the left-hand side.
$ \Rightarrow \left( {\dfrac{{4 + 2y + 12y}}{3}} \right) = 3$
That is equal to,
$ \Rightarrow \left( {\dfrac{{4 + 14y}}{3}} \right) = 3$
Multiply by 3 on both sides.
$ \Rightarrow \left( {\dfrac{{4 + 14y}}{3}} \right) \times 3 = 3 \times 3$
That is
$ \Rightarrow 4 + 14y = 9$
Now, subtract 4 on both sides.
$ \Rightarrow 4 - 4 + 14y = 9 - 4$
That is equal to,
$ \Rightarrow 14y = 5$
Now, divide by 14 into both sides.
$ \Rightarrow \dfrac{{14y}}{{14}} = \dfrac{5}{{14}}$
So,
$ \Rightarrow y = \dfrac{5}{{14}}$
Now, put the value of y in equation (1).
$ \Rightarrow 3x - 2y = 4$
Put the value of y.
$ \Rightarrow 3x - 2\left( {\dfrac{5}{{14}}} \right) = 4$
That is equal to
$ \Rightarrow 3x - \left( {\dfrac{5}{7}} \right) = 4$
Let us add $\dfrac{5}{7}$ to both sides.
 $ \Rightarrow 3x - \left( {\dfrac{5}{7}} \right) + \left( {\dfrac{5}{7}} \right) = 4 + \left( {\dfrac{5}{7}} \right)$
So,
$ \Rightarrow 3x = 4 + \left( {\dfrac{5}{7}} \right)$
Let us take LCM on the right-hand side.
 $ \Rightarrow 3x = \left( {\dfrac{{28 + 5}}{7}} \right)$
That is
$ \Rightarrow 3x = \left( {\dfrac{{33}}{7}} \right)$
Now, divide both sides by 3.
$ \Rightarrow \dfrac{{3x}}{3} = \left( {\dfrac{{33}}{{7 \times 3}}} \right)$
So,
$ \Rightarrow x = \dfrac{{11}}{7}$
Hence, we find the value of x is $\dfrac{{11}}{7}$ and the value of y is $\dfrac{5}{{14}}$ using substitution methods.

Note: To check 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 - 2y = 4$
Put $x = \dfrac{{11}}{7}$ and $y = \dfrac{5}{{14}}$.
$ \Rightarrow 3\left( {\dfrac{{11}}{7}} \right) - 2\left( {\dfrac{5}{{14}}} \right) = 4$
$ \Rightarrow \dfrac{{33}}{7} - \dfrac{5}{7} = 4$
 $ \Rightarrow \dfrac{{28}}{7} = 4$
That is equal to,
$ \Rightarrow 4 = 4$
Now, let us take equation (2) and put the values.
 $ \Rightarrow x + 4y = 3$
Put $x = \dfrac{{11}}{7}$ and $y = \dfrac{5}{{14}}$.
$ \Rightarrow \left( {\dfrac{{11}}{7}} \right) + 4\left( {\dfrac{5}{{14}}} \right) = 3$
That is
$ \Rightarrow \left( {\dfrac{{11}}{7}} \right) + \left( {\dfrac{{10}}{7}} \right) = 3$
$ \Rightarrow \left( {\dfrac{{21}}{7}} \right) = 3$
That is equal to,
$ \Rightarrow 3 = 3$