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# How do you solve the system $0.3x-0.2y=0.5$ and $x-2y=-5$ using substitution?

Last updated date: 13th Jun 2024
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Hint: From the question given, we have been asked to solve the system $0.3x-0.2y=0.5$ and $x-2y=-5$ using substitution. We can solve the above given equations by using the process of substitution. By using the substitution method, first we will get one variable value and by using that we have to find another variable value.

From the question given, it has been given that $0.3x-0.2y=0.5$
Let us assume this equation be $\left( 1 \right)$.
$x-2y=-5$
Let us assume this equation be $\left( 2 \right)$.
First of all, let us solve the second equation,
$x-2y=-5$
Shift $-2y$ from left hand side of the equation to the right hand side of the equation.
By shifting $-2y$ from left hand side of the equation to the right hand side of the equation, we get
$\Rightarrow x=2y-5$
Now, let us assume this equation be $\left( 3 \right)$.
Now, let us substitute this equation $\left( 3 \right)$ into the first equation to get the value of $y$.
By substituting, we get
$\Rightarrow 0.3x-0.2y=0.5$
$\Rightarrow 0.3\left( 2y-5 \right)-0.2y=0.5$
Simplify more further to get the exact value. By simplifying further, we get
$\Rightarrow 0.6y-1.5-0.2y=0.5$
$\Rightarrow 0.4y-1.5=0.5$
Shift $-1.5$ to the right hand side of the equation. By shifting, we get
$\Rightarrow 0.4y=2$
$\Rightarrow y=\dfrac{2}{0.4}$
$\Rightarrow y=\dfrac{20}{4}$
$\Rightarrow y=5$
Now, substitute $y=5$ in the equation $\left( 3 \right)$. By substituting we get,
$x=2y-5$
$\Rightarrow x=2\left( 5 \right)-5$
$\Rightarrow x=10-5$
$\Rightarrow x=5$

Hence, the solution is $x=5\text{ and y=5}$.

Note: We should be very careful while doing the calculation in this problem. Also, we should know all methods to solve the given equations. For this question we have chosen a substitution method. If we are not satisfied with the obtained values then they can verify the values by substituting them in the given system of equations. Calculation should be done very carefully while finding the solution for the given question. Similarly we can solve the system $3x-2y-5=0$ and $x-2y-5=0$ given as $x=2y+5$ by substituting it in the other equation we will have $3\left( 2y+5 \right)-2y-5=0\Rightarrow 6y+15-2y-5=0\Rightarrow 4y+10=0\Rightarrow y=-\dfrac{5}{2}$ and we will have $x=0$