Question

How do you solve $x - 4y - 1 = 0$ and $x + 5y - 4 = 0$ using substitution?

Answer 1

Hint: In this question, we need to solve the given equations by the method substitution. Begin with the first equation. Make rearrangement in the first equation to get the expression for the unknown variable x. Now consider the second equation. In the second equation replace the variable x by the expression obtained for x from the first equation. Thus, we obtain an equation in only one variable which is y. Now solve the equation obtained, to get the value of y. Then substitute back any one of the equations to get the value of x. So the required solution is obtained.

Complete step by step answer:
Given two equations of the form,
$x - 4y - 1 = 0$ …… (1)
$x + 5y - 4 = 0$ …… (2)
We are asked to solve the above equations by the method of substitution.
i.e. we find the values of the unknown variable x and y by the substitution method.
Now let us consider the first equation. We make rearrangement and simplify it to obtain the expression for x.
By equation (1), we have,
$x - 4y - 1 = 0$
Transfer the constant term to the R.H.S. we get,
$ \Rightarrow x - 4y = 1$
Now transfer $4y$ to the R.H.S. we get,
$ \Rightarrow x = 4y + 1$ …… (3)
Hence we obtained an expression for the variable x.
Remember that when transferring numbers or variables to the other side, the signs of the same will be changed to the opposite sign.
Let us consider the second equation.
From the equation (2), we have,
$x + 5y - 4 = 0$
Now substitute the expression of x given by the equation (3) and obtain an expression in only one variable which is y.
Substituting equation (3) in $x + 5y - 4 = 0$, we get,
$ \Rightarrow 4y + 1 + 5y - 4 = 0$
Rearranging the terms we get,
$ \Rightarrow 4y + 5y + 1 - 4 = 0$
Combining the like terms we get,
$ \Rightarrow 9y - 3 = 0$
Taking 3 to the other side we get,
$ \Rightarrow 9y = 3$
Dividing by 9 on both sides, we get,
$ \Rightarrow \dfrac{{9y}}{9} = \dfrac{3}{9}$
$ \Rightarrow y = \dfrac{1}{3}$
Substitute y value in the equation (3), we get,
$ \Rightarrow x = 4\left( {\dfrac{1}{3}} \right) + 1$
$ \Rightarrow x = \dfrac{4}{3} + 1$
Taking LCM we get,
$ \Rightarrow x = \dfrac{{4 + 3}}{3}$
$ \Rightarrow x = \dfrac{7}{3}$

Hence, the solution for the given equations by substitution method is given by $x = \dfrac{7}{3}$ and $y = \dfrac{1}{3}$.

Note: We can check whether our solution is correct or not, by substituting the obtained values of the variables x and y in any one of the equations. If the L.H.S. and R.H.S. of the equation are equal, then our answer is correct. Otherwise we are wrong.
It is important to know the following basic facts.
An equation remains unchanged or undisturbed if it satisfies the following conditions.
(1) If L.H.S. and R.H.S. are interchanged.
(2) If the same number is added on both sides of the equation.
(3) If the same number is subtracted on both sides of the equation.
(4) When both L.H.S. and R.H.S. are multiplied by the same number.
(5) When both L.H.S. and R.H.S. are divided by the same number.