Question

How do you solve \[\dfrac{1}{5}x + \dfrac{1}{3}y = \dfrac{{ - 1}}{{15}}\;\] and \[\dfrac{1}{5}x + 2y = \dfrac{8}{5}\] using substitution?

Answer 1

Hint: In this question, we have been given two linear equations in which we need to solve for values of variables using the substitution method. We can solve for one variable, and then substitute that expression into the other equation. The important thing here is that you are always substituting values that are equivalent.

Complete step-by-step solution:
\[ \Rightarrow \dfrac{1}{5}x + \dfrac{1}{3}y = \dfrac{{ - 1}}{{15}}\] …….…. (A), and
\[ \Rightarrow \dfrac{1}{5}x + 2y = \dfrac{8}{5}\;\] …….…. (B)
The motive of the substitution method is to write one of the equations in terms of a single variable- either x or y. Equation B tells us that, \[ \Rightarrow \dfrac{1}{5}x + 2y = \dfrac{8}{5}\;\] so it makes sense to substitute that \[\dfrac{8}{5} - 2y\] in place of \[\dfrac{1}{5}x\] in Equation A.
Substituting \[\dfrac{8}{5} - 2y\] in place of \[\dfrac{1}{5}x\] into Equation A, we will get the following equation,
\[\dfrac{8}{5} - 2y + \dfrac{1}{3}y = \dfrac{{ - 1}}{{15}}\] ……….. (C)
Simplify and solve the equation(C) for y.
\[\dfrac{8}{5} + \dfrac{1}{{15}} = 2y - \dfrac{1}{3}y\]
Simplifying the above equation further by taking lcm of the terms in the denominator we get:
\[\dfrac{{24 + 1}}{{15}} = 2y - \dfrac{1}{3}y\] …. (D)
Further solving equation (D) we will get the value of y as shown:
\[\dfrac{{25}}{{15}} = \dfrac{5}{3}y\;\]
\[y = 1\] …. (1)
To now find x, substitute this value for y from equation (1) into either equation and solve for x. We will use Equation A here by putting \[y = 1\] to get the value of x.
\[\dfrac{1}{5}x + \dfrac{1}{3} \times {\text{1}} = \dfrac{{ - 1}}{{15}}\]
Simplifying further to find the value of x, we will get,
\[\dfrac{1}{5}x = \dfrac{{ - 1}}{{15}}\; - \dfrac{1}{3} = \dfrac{{ - 1 - 5}}{{15}} = \dfrac{{ - 6}}{{15}} = \dfrac{{ - 2}}{5}\]
\[\dfrac{1}{5}x{\text{ }} = \dfrac{{ - 2}}{5}\]
Thus, we get the value of \[x = - 2\] from the above calculation using the substitution method.
Finally, check the solution we get \[x = - 2\], \[y = 1\] by substituting these values into each of the original equations.

Hence the solution is \[\left( { - 2,{\text{ }}1} \right)\].

Note: We can also use the addition method to derive the conclusion instead of the substitution method. Remember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair \[\left( { - 2,{\text{ }}1} \right)\] does work for both equations, so you know that it is a solution to the system as well. Sometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute.