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

How do you solve the system 2x+y=5,3x−y=102x + y = 5,3x - y = 10 ?

Explanation

Solution

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=52x + y = 5 and 3x−y=103x - y = 10
Let us solve the first equation to obtain the value of x in terms of y.
We have 2x+y=52x + y = 5
Shift the value of y to RHS of the equation.
⇒2x=5−y\Rightarrow 2x = 5 - y
Divide both sides by 2
⇒x=5−y2\Rightarrow x = \dfrac{{5 - y}}{2} … (1)
Now we substitute the value of x=5−y2x = \dfrac{{5 - y}}{2} from equation (1) in the second linear equation.
Substitute x=5−y2x = \dfrac{{5 - y}}{2}in 3x−y=103x - y = 10
⇒3(5−y2)−y=10\Rightarrow 3(\dfrac{{5 - y}}{2}) - y = 10
Open the bracket in LHS of the equation
⇒15−3y2−y=10\Rightarrow \dfrac{{15 - 3y}}{2} - y = 10
Take LCM in left hand side of the equation
⇒15−3y−2y2=10\Rightarrow \dfrac{{15 - 3y - 2y}}{2} = 10
Add like terms in numerator of LHS of the equation
⇒15−5y2=10\Rightarrow \dfrac{{15 - 5y}}{2} = 10
Take 5 common from numerator of LHS
⇒5(3−y)2=5×2\Rightarrow \dfrac{{5(3 - y)}}{2} = 5 \times 2
Cancel same factors from both sides of the equation
⇒3−y2=2\Rightarrow \dfrac{{3 - y}}{2} = 2
Cross multiply the denominator from LHS to RHS of the equation
⇒3−y=4\Rightarrow 3 - y = 4
Shift the constant values to one side of the equation.
⇒−y=4−3\Rightarrow - y = 4 - 3
⇒−y=1\Rightarrow - y = 1
Divide both sides by -1
⇒−y−1=1−1\Rightarrow \dfrac{{ - y}}{{ - 1}} = \dfrac{1}{{ - 1}}
Cancel the same terms from numerator and denominator.
⇒y=−1\Rightarrow y = - 1
Now substitute the value of y=−1y = - 1 in equation (1) to get the value of x.
⇒x=5−(−1)2\Rightarrow x = \dfrac{{5 - ( - 1)}}{2}
Open the bracket in RHS of the equation.
⇒x=5+12\Rightarrow x = \dfrac{{5 + 1}}{2}
⇒x=62\Rightarrow x = \dfrac{6}{2}
Cancel the same terms from numerator and denominator.
⇒x=3\Rightarrow x = 3
So, the value of x is 3.

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

Note:
Alternate method:
We are given the equations 2x+y=52x + y = 5 and 3x−y=103x - y = 10
We find the value of y in terms of x from the first equation
⇒2x+y=5\Rightarrow 2x + y = 5
Shift the value of y to RHS of the equation.
⇒y=5−2x\Rightarrow y = 5 - 2x … (2)
Substitute the value of x in equation 3x−y=103x - y = 10
⇒3x−(5−2x)=10\Rightarrow 3x - (5 - 2x) = 10
Open the bracket in LHS of the equation
⇒3x−5+2x=10\Rightarrow 3x - 5 + 2x = 10
Shift constant values to one side of the equation.
⇒5x=10+5\Rightarrow 5x = 10 + 5
⇒5x=15\Rightarrow 5x = 15
Divide both sides by 5
⇒5x5=155\Rightarrow \dfrac{{5x}}{5} = \dfrac{{15}}{5}
Cancel the same terms from numerator and denominator.
⇒x=3\Rightarrow x = 3
So, the value of x is 3
Put in equation (2) and find y
⇒y=5−2×3\Rightarrow y = 5 - 2 \times 3
⇒y=5−6\Rightarrow y = 5 - 6
⇒y=−1\Rightarrow y = - 1
∴\therefore Solution of the system of linear equations is x=3;y=−1x = 3;y = - 1