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How do you solve $5x-3y=-20$ and $-3x+6y=12$?

Last updated date: 24th Feb 2024
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IVSAT 2024
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Hint: There are two unknowns $x$ and $y$ and also two equations to solve. We solve the equations equating the coefficients of one variable and omitting the variable. The other variable remains with the constants. Using the binary operation, we find the value of the other variable. First, we are applying the process of reduction and then the substitution.

Complete step-by-step solution:
The given equations $5x-3y=-20$ and $-3x+6y=12$ are linear equations of two variables.
We know that the number of equations has to be equal to the number of unknowns to solve them.
We take the equations as $5x-3y=-20.....(i)$ and $-3x+6y=12......(ii)$.
We multiply 2 to the both sides of the first equation and get
  & 2\times \left( 5x-3y \right)=\left( -20 \right)\times 2 \\
 & \Rightarrow 10x-6y=-40 \\
We take the equation as $10x-6y=-40.....(iii)$.
Now we add the equation (ii) from equation (iii) and get
$\left( 10x-6y \right)+\left( -3x+6y \right)=-40+12$.
We take the variables together and the constants on the other side.
Simplifying the equation, we get
  & \left( 10x-6y \right)+\left( -3x+6y \right)=-40+12 \\
 & \Rightarrow 10x-3x=-28 \\
 & \Rightarrow 7x=-28 \\
 & \Rightarrow x=\dfrac{-28}{7}=-4 \\
The value of $x$ is $-4$. Now putting the value in the equation $5x-3y=-20.....(i)$, we get
  & 5x-3y=-20 \\
 & \Rightarrow y=\dfrac{5x+20}{3}=\dfrac{5\times \left( -4 \right)+20}{3} \\
 & \Rightarrow y=\dfrac{-20+20}{3}=0 \\
Therefore, the values are $x=-4,y=0$.

Note: We can also find the value of one variable $y$ with respect to $x$ based on the equation
$5x-3y=-20$ where $y=\dfrac{5x+20}{3}$. We replace the value of $y$ in the second equation of
$-3x+6y=12$ and get
  & -3x+6y=12 \\
 & \Rightarrow -3x+6\left( \dfrac{5x+20}{3} \right)=12 \\
 & \Rightarrow -9x+30x+120=36 \\
 & \Rightarrow 21x=36-120=-84 \\
We get the equation of $x$ and solve
  & 21x=-84 \\
 & \Rightarrow x=\dfrac{-84}{21}=-4 \\
Putting the value of $x$ we get $y=\dfrac{5x+20}{3}=\dfrac{5\left( -4 \right)+20}{3}=0$.
Therefore, the values are $x=-4,y=0$.

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