Question

How do you determine whether a linear system has one solution, many solutions, or no solution when given $5x+4y=-18$ and $2x+3y=-24$?

Answer 1

Hint: We find the general equations and their coefficients to find ratios. We define the relations between those ratios to find the number of solutions possible. Then we put the values from $5x+4y=-18$ and $2x+3y=-24$ to find a number of solutions.

Complete step-by-step solution:
The coefficients of the given equation define the number of solutions for the given equations.
Let’s take two equations and two unknowns. $ax+by=m$ and $cx+dy=n$
In the equations the terms a, b, c, d, m, n all are constants.
We take the ratios of the coefficients of the same variables. The ratios are $\dfrac{a}{c},\dfrac{b}{d},\dfrac{m}{n}$.
The relation among these ratios determines the equations.
If $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{m}{n}$, then we have an infinite number of solutions for those equations.
If $\dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n}$, then we have no solution for those equations.
If $\dfrac{a}{c}\ne \dfrac{b}{d}$, then we have only one solution for those equations.
In case of $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{m}{n}$, both equations become a single equation. For a particular value of x, we can determine the value of y.
In case of $\dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n}$, for particular values of x and y, we have different values of same function $ax+by=cx+dy$.
Now, for $5x+4y=-18$ and $2x+3y=-24$. We take the coefficients and place them in the formula. The ratios are $\dfrac{5}{2},\dfrac{4}{3},\dfrac{-18}{-24}$. We have $\dfrac{-18}{-24}=\dfrac{3}{4}$.
The relation is $\dfrac{5}{2}\ne \dfrac{4}{3}\ne \dfrac{-18}{-24}$. So, we have one solution to those equations.

Note: The solution of $5x+4y=-18$ and $2x+3y=-24$. Adding them we get
$\begin{align}
  & \left( 5x+4y \right)+\left( 2x+3y \right)=-18-24 \\
 & \Rightarrow 7x+7y=-42 \\
 & \Rightarrow 4x+4y=\dfrac{-42\times 4}{7}=-24 \\
\end{align}$
Now we take subtraction and get
$\begin{align}
  & \Rightarrow \left( 5x+4y \right)-\left( 4x+4y \right)=-18-\left( -24 \right) \\
 & \Rightarrow x=6 \\
\end{align}$
Putting the value in equation we get $y=\dfrac{-18-5x}{4}=\dfrac{-18-5\times 6}{4}=\dfrac{-48}{4}=-12$.
The solution is $x=6,y=-12$.