Question

How do you solve $y=-3x+5$ and $5x-4y=-3$ using substitution? \[\]

Answer 1

Hint: 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 $x$ from one of the expression and put $y$ in terms of $x$ in other equation to get the value of $x$. We are already given the first equation $y=-3x+5$ in terms of $x$. We put $y$ in the second equation and solve for $x$ and then put $x$ in first equation to get $y$.\[\]

Complete step-by-step answer:
We know that the general linear equation is given by $ax+by+c=0$ where $a,b,c$are real numbers and $a\ne 0,b\ne 0$ .We need at least two equations to find a unique solution. We are given the following pair of equations in the equation in the question.
\[\begin{align}
  & y=-3x+5.....\left( 1 \right) \\
 & 5x-4y=-3....\left( 2 \right) \\
\end{align}\]
We know from substation methods to solve linear equations that we have to express $y$ in terms of $x$ from one of the equations and then put $y$ in the other equation. We see that the first equation is already given to us $y$ as an expression of $x$. So we put $y=3x+5$ in equation (2) to have
\[\begin{align}
  & \Rightarrow 5x-4\left( -3x+5 \right)=-3 \\
 & \Rightarrow 5x+12x-20=-3 \\
 & \Rightarrow 17x-20=-3 \\
\end{align}\]
 We see the above expression is now a linear equation with only one variable that is $x$. We add 20 both sides of above equation to have
\[\begin{align}
  & \Rightarrow 17x-20+20=-3+20 \\
 & \Rightarrow 17x=17 \\
\end{align}\]
We divide both sides of the above step by 17 to have
\[\Rightarrow x=1\]
We put obtained value of $x=1$ in equation (1) to have
\[y=-3\times 1+5=-3+5=2\]
So the solution of the given equations is $x=1,y=2$.\[\]

Note: We note that if ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ be two linear equations then we can oblation unique solution only when ratio between coefficients of variables is not equal that is $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$. Here in this problem we have the ratio of coefficients $\dfrac{-3}{5}\ne \dfrac{-1}{-3}$.We should check this in rough before solving the equation. We can substitute $x$ as an expression of $y$ to alternatively solve.