Question

How do you solve the system \[2x + y = 5,3x - y = 10\] ?

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 \[2x + y = 5\] and \[3x - y = 10\]
Let us solve the first equation to obtain the value of x in terms of y.
We have \[2x + y = 5\]
Shift the value of y to RHS of the equation.
\[ \Rightarrow 2x = 5 - y\]
Divide both sides by 2
\[ \Rightarrow x = \dfrac{{5 - y}}{2}\] … (1)
Now we substitute the value of \[x = \dfrac{{5 - y}}{2}\] from equation (1) in the second linear equation.
Substitute \[x = \dfrac{{5 - y}}{2}\]in \[3x - y = 10\]
\[ \Rightarrow 3(\dfrac{{5 - y}}{2}) - y = 10\]
Open the bracket in LHS of the equation
\[ \Rightarrow \dfrac{{15 - 3y}}{2} - y = 10\]
Take LCM in left hand side of the equation
\[ \Rightarrow \dfrac{{15 - 3y - 2y}}{2} = 10\]
Add like terms in numerator of LHS of the equation
\[ \Rightarrow \dfrac{{15 - 5y}}{2} = 10\]
Take 5 common from numerator of LHS
\[ \Rightarrow \dfrac{{5(3 - y)}}{2} = 5 \times 2\]
Cancel same factors from both sides of the equation
\[ \Rightarrow \dfrac{{3 - y}}{2} = 2\]
Cross multiply the denominator from LHS to RHS of the equation
\[ \Rightarrow 3 - y = 4\]
Shift the constant values to one side of the equation.
\[ \Rightarrow - y = 4 - 3\]
\[ \Rightarrow - y = 1\]
Divide both sides by -1
\[ \Rightarrow \dfrac{{ - y}}{{ - 1}} = \dfrac{1}{{ - 1}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow y = - 1\]
Now substitute the value of \[y = - 1\] in equation (1) to get the value of x.
\[ \Rightarrow x = \dfrac{{5 - ( - 1)}}{2}\]
Open the bracket in RHS of the equation.
\[ \Rightarrow x = \dfrac{{5 + 1}}{2}\]
\[ \Rightarrow x = \dfrac{6}{2}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow x = 3\]
So, the value of x is 3.

\[\therefore \]Solution of the system of linear equations is \[x = 3;y = - 1\]

Note:
Alternate method:
We are given the equations \[2x + y = 5\] and \[3x - y = 10\]
We find the value of y in terms of x from the first equation
\[ \Rightarrow 2x + y = 5\]
Shift the value of y to RHS of the equation.
\[ \Rightarrow y = 5 - 2x\] … (2)
Substitute the value of x in equation \[3x - y = 10\]
\[ \Rightarrow 3x - (5 - 2x) = 10\]
Open the bracket in LHS of the equation
\[ \Rightarrow 3x - 5 + 2x = 10\]
Shift constant values to one side of the equation.
\[ \Rightarrow 5x = 10 + 5\]
\[ \Rightarrow 5x = 15\]
Divide both sides by 5
\[ \Rightarrow \dfrac{{5x}}{5} = \dfrac{{15}}{5}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow x = 3\]
So, the value of x is 3
Put in equation (2) and find y
\[ \Rightarrow y = 5 - 2 \times 3\]
\[ \Rightarrow y = 5 - 6\]
\[ \Rightarrow y = - 1\]
\[\therefore \]Solution of the system of linear equations is \[x = 3;y = - 1\]