Question

How do you solve the linear system of equations $x+3y=5$ and $2x-y=5$ ?

Answer 1

We recall that from substitution method that if we are given two linear equation ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{ 2}}y+{{c}_{2}}=0$ then we express $y$ in terms of $x$ (or $x$ in terms of $y$)from one of the expression and put $y$ in terms of $x$ in other equation to get the value of $x$. We then put the value of $x$ in either equation to get the value of $y$.

Complete step by step solution:
We are given the following pair of equations in the equation in the question.

& x+3y=5.....\left( 1 \right) \\\ & 2x-y=5....\left( 2 \right) \\\ \end{aligned}$$ We know from the substitution method to solve linear equations that we have to express $y$ in terms of $x$ (or $x$ in terms of $y$) from one of the equations and then put $y$ in the other equation. We express $y$ in terms $x$ from the second equation as $$\begin{aligned} & 2x-y=5 \\\ & \Rightarrow -y=5-2x \\\ & \Rightarrow y=2x-5........\left( 3 \right) \\\ \end{aligned}$$ We put above expression of $y$ in terms of $x$ to in equation (1) have; $$\begin{aligned} & x+3\left( 2x-5 \right)=5 \\\ & \Rightarrow x+6x-15=5 \\\ & \Rightarrow 7x-15=5 \\\ \end{aligned}$$ We see the above expression is now a linear equation only in one variable that is $x$. We add 15 both sides of above equation to have $$\begin{aligned} & \Rightarrow 7x=20 \\\ & \Rightarrow x=\dfrac{20}{7} \\\ \end{aligned}$$ We put obtained value of $x=\dfrac{20}{7}$ in equation (3) to have $$y=2\left( \dfrac{20}{7} \right)-5=\dfrac{40}{7}-5=\dfrac{40-35}{7}=\dfrac{5}{7}$$ So the solution of the given equations is $x=\dfrac{20}{7},y=\dfrac{5}{7}$.$$$$ **Note:** We note that two linear equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{ 2}}y+{{c}_{2}}=0$ have unique solution when $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$. We should check this ratio every time we are asked to solve linear equations. We can alternately solve by elimination method where we eliminate either of the variables. We eliminate $y$ from the pair of equations by multiplying 3 to equation and then adding both equations side by side as $$\begin{aligned} & x+3y+3\left( 2x-y \right)=5+3\times 5 \\\ & \Rightarrow x+3y+6x-3y=5+15 \\\ & \Rightarrow 7x=20 \\\ & \Rightarrow x=\dfrac{20}{7} \\\ \end{aligned}$$ We put the above values in equation (2) to get $y=\dfrac{5}{7}$. We can quickly solve using the formula $x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}\,},y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}\,}$.