Question

How do you solve $y=6x$ and $2x+3y=-20$ 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 $y=6x$ 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.
\[\begin{align}
  & y=6x.....\left( 1 \right) \\
 & 2x+3y=-20....\left( 2 \right) \\
\end{align}\]
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 equation (1) is already given to us $y$ as an expression of $x$. So we put $y=6x$ in equation (2) to have
\[\begin{align}
  & \Rightarrow 2x+3\left( 6x \right)=-20 \\
 & \Rightarrow 2x+18x=-20 \\
 & \Rightarrow 20x=-20 \\
\end{align}\]
 We see the above expression is now a linear equation only in one variable that is $x$. We divide both sides of the above equation by 20 to have
\[\Rightarrow x=-1\]
We put obtained value of $x=-1$ in equation (1) to have
\[y=6\left( -1 \right)=-6\]
So the solution of the given equations is $x=-1,y=-6$\[\]

Note: We can put $x=-1,y=6$ in the left hand side of second equation to have$2\left( -1 \right)+3\left( -6 \right)=-20$. Hence our solution is verified. 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 a ratio of coefficients $\dfrac{6}{2}\ne \dfrac{-1}{3}$.We should check this in rough before solving the equation.