Question

How do you solve by substitution \[\dfrac{1}{2}x + 2y = 12\] and \[x - 2y = 6\] ?

Answer 1

Hint: We use the substitution method to solve two linear equations given in the question. We find the value of x from the first equation in terms of y and substitute in the second equation which becomes an equation in y entirely. Solve for the value of y and substitute back the value of y to obtain the value of x.

Complete step-by-step answer:
We have two linear equations \[\dfrac{1}{2}x + 2y = 12\] and \[x - 2y = 6\]
Let us solve the first equation to obtain the value of x in terms of y.
We have \[\dfrac{1}{2}x + 2y = 12\]
Shift the value of y to the right hand side of the equation.
\[ \Rightarrow \;\dfrac{1}{2}x = 12 - 2y\]
Multiply both sides by 2
\[ \Rightarrow x = (12 - 2y) \times 2\]
\[ \Rightarrow x = 24 - 4y\] … (1)
Now we substitute the value of \[x = 24 - 4y\] from equation (1) in the second linear equation.
Substitute \[x = 24 - 4y\] in \[x - 2y = 6\]
\[ \Rightarrow (24 - 4y) - 2y = 6\]
Open the bracket in left hand side of the equation
\[ \Rightarrow 24 - 4y - 2y = 6\]
Add or subtract like terms in left hand side of the equation
\[ \Rightarrow 24 - 6y = 6\]
Cancel same factors from both sides of the equation i.e. 6
\[ \Rightarrow 4 - y = 1\]
Shift y to right hand side of the equation
\[ \Rightarrow 4 - 1 = y\]
\[ \Rightarrow y = 3\]
Now substitute the value of \[y = 3\] in equation (1) to get the value of x.
\[ \Rightarrow x = 24 - 4 \times 3\]
Calculate the product on right hand side of the equation
\[ \Rightarrow x = 24 - 12\]
Calculate the difference on right hand side of the equation
\[ \Rightarrow x = 12\]
So, the value of x is 12

\[\therefore \] Solution of the system of linear equations using substitution method is \[x = 12;y = 3\]

Note:
Students many times make the mistake of not changing the sign when shifting the values from one side of the equation to another side, always keep in mind the sign changes from positive to negative and vice versa when we shift a number from one side to another side of the equation.