Question

How do you solve the following system using substitution: \[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}\], \[\dfrac{3}{x} - \dfrac{1}{y} = \dfrac{1}{4}\]?

Answer 1

Hint: Here in this question, given the system of linear equations. We have to find the unknown values that are \[x\] and \[y\] by using the substitution method. First Solve 1 equation for 1 variable. (Put in y = or x = form) Substitute this expression into the other equation and solve for the missing variable.

Complete step-by-step solution:
The substitution method is a method of solving systems of linear equations. The main idea behind solving systems with the substitution method is to choose one of the equations, solve it for one of the variables, and plug the result into the other equation. This way, we obtain an equation with one variable, which we can easily solve.
Let us consider the given system of linear equations
\[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}\]----------(1)
\[\dfrac{3}{x} - \dfrac{1}{y} = \dfrac{1}{4}\]----------(2)
Solve one of the equations for either x = or y =.
Consider equation (1)
\[ \Rightarrow \,\,\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}\]
Keep on left hand side remaining term should move to the RHS
\[ \Rightarrow \,\,\dfrac{1}{y} = \dfrac{3}{4} - \dfrac{1}{x}\]--------(3)
Substitute the solution from equation (3) into the other equation (2).
\[ \Rightarrow \,\,\,\dfrac{3}{x} - \left( {\dfrac{3}{4} - \dfrac{1}{x}} \right) = \dfrac{1}{4}\]
Remove parenthesis by multiplying ‘ – ve‘ sign
\[ \Rightarrow \,\,\,\dfrac{3}{x} - \dfrac{3}{4} + \dfrac{1}{x} = \dfrac{1}{4}\]
On simplification, we get
 \[ \Rightarrow \,\,\,\dfrac{1}{x} = \dfrac{1}{4}\]
Take reciprocal
\[\therefore \,\,\,\,x = 4\]
We have found the value of \[x\] now we have to find the value of \[y\] . so we will substitute the value \[x\]to any one of the equation (1) or (2) . we will substitute the value of \[x\]to equation (1).
Therefore, we have \[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}\]
\[ \Rightarrow \,\,\dfrac{1}{4} + \dfrac{1}{y} = \dfrac{3}{4}\]
Subtract \[\dfrac{1}{4}\] on both side, then
\[ \Rightarrow \,\,\dfrac{1}{4} + \dfrac{1}{y} - \dfrac{1}{4} = \dfrac{3}{4} - \dfrac{1}{4}\]
\[ \Rightarrow \,\,\dfrac{1}{y} = \dfrac{3}{4} - \dfrac{1}{4}\]
\[ \Rightarrow \,\,\dfrac{1}{y} = \dfrac{{3 - 1}}{4}\]
\[ \Rightarrow \,\,\dfrac{1}{y} = \dfrac{2}{4}\]
\[ \Rightarrow \,\,\dfrac{1}{y} = \dfrac{1}{2}\]
Take reciprocal
\[\therefore \,\,\,\,y = 2\]
Hence, we got the unknown values \[x\] and \[y\]that is \[4\] and \[2\] respectively,
We can check whether these values are correct or not by substituting the unknown values in the given equations and we have to prove L.H.S is equal to R.H.S
Now we will substitute the value of \[x\] and \[y\] in equation (1) so we have
\[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}\]
\[ \Rightarrow \,\,\,\dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}\]
Take LCM 4 and 2 is 4
\[ \Rightarrow \,\,\,\dfrac{{1 + 2}}{4} = \dfrac{3}{4}\]
\[ \Rightarrow \,\,\,\dfrac{3}{4} = \dfrac{3}{4}\]
\[\therefore \,\,\,LHS = RHS\]
Hence the values of the unknown that are \[x\] and \[y\] are the correct values which satisfy the equation.

Note: In this type of question while substituting the term we must be aware of the sign where we change the sign by the alternate sign. In this we have a chance to verify our answers. In the substitution method we have made the one term have the same coefficient such that it will be easy to solve the equation.