Question

How do you solve \[0.3x-0.2y=4\] and \[0.4x+0.5y=\dfrac{63}{19}\] using substitution?

Answer 1

Hint: This question is from the topic of algebra. We are going to solve these equations using a substitution method. In this question, we are going to find out the value of x and y. We will find the value of x from the equation \[0.3x-0.2y=4\] and put that value of x in the equation \[0.4x+0.5y=\dfrac{63}{19}\]. From there, we will find the value of y. And, after using any one of the equations we will find the exact value of x.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to solve the equations \[0.3x-0.2y=4\] and \[0.4x+0.5y=\dfrac{63}{19}\] & find out the value of x and y.
The equation \[0.3x-0.2y=4\] can also be written as
\[\Rightarrow 0.3x=4+0.2y\]
The above equation can also be written as
\[\Rightarrow x=\dfrac{4+0.2y}{0.3}\]
The above equation can also be written as
\[\Rightarrow x=\dfrac{4}{0.3}+\dfrac{0.2y}{0.3}\]
As we know that 0.2 can also be written as 2 divided by 10 and 0.3 can also be written as 3 divided by 10. So, we can write the above equation as
\[\Rightarrow x=\dfrac{4}{\dfrac{3}{10}}+\dfrac{2y}{10\times \dfrac{3}{10}}\]
The above equation can also be written as
\[\Rightarrow x=\dfrac{40}{3}+\dfrac{2y}{3}\]
Now, we will put this value of x in the second equation that is \[0.4x+0.5y=\dfrac{63}{19}\].
\[0.4\left( \dfrac{40}{3}+\dfrac{2y}{3} \right)+0.5y=\dfrac{63}{19}\]
The above equation can also be written as
\[\Rightarrow 0.4\left( \dfrac{40}{3} \right)+0.4\left( \dfrac{2y}{3} \right)+0.5y=\dfrac{63}{19}\]
As we know that if we multiply 0.4 with 40, then we will get 16.
So, we can write the above equation as
\[\Rightarrow \dfrac{16}{3}+\dfrac{0.8}{3}y+0.5y=\dfrac{63}{19}\]
The above equation can also be written as
\[\Rightarrow \dfrac{16}{3}+\left( \dfrac{0.8+1.5}{3} \right)y=\dfrac{63}{19}\]
Taking the term \[\dfrac{16}{3}\] to the right side of the equation, we get
\[\Rightarrow \left( \dfrac{2.3}{3} \right)y=\dfrac{63}{19}-\dfrac{16}{3}\]
The above equation can also be written as
\[\Rightarrow \left( \dfrac{2.3}{3} \right)y=\dfrac{63\times 3-16\times 19}{19\times 3}\]
The above can also be written as
\[\Rightarrow \left( \dfrac{2.3}{3} \right)y=\dfrac{189-304}{19\times 3}=\dfrac{-115}{19\times 3}\]
The above equation can also be written as
\[\Rightarrow \left( 2.3 \right)y=\dfrac{-115}{19}\]
As we know that 2.3 can also be written as 23 divided by 10, so we can write the above equation as
\[\Rightarrow \dfrac{23}{10}y=\dfrac{-115}{19}\]
We know that, 115 can also be written as 23 multiplied by 5. So, we can remove 23 to both the side of the equation, then we will get
\[\Rightarrow \dfrac{1}{10}y=\dfrac{-5}{19}\]
The above can also be written as
\[\Rightarrow y=\dfrac{-50}{19}\]
Putting this value of y in the equation \[0.3x-0.2y=4\], we get
\[0.3x-0.2\left( \dfrac{-50}{19} \right)=4\]
The above equation can also be written as
\[\Rightarrow 0.3x+\dfrac{10}{19}=4\]
The above equation can also be written as
\[\Rightarrow 0.3x=4-\dfrac{10}{19}\]
\[\Rightarrow 0.3x=\dfrac{76-10}{19}=\dfrac{66}{19}\]
The above equation can also be written as
\[\Rightarrow \dfrac{3}{10}x=\dfrac{66}{19}\]
The above equation can also be written as
\[\Rightarrow \dfrac{3}{10}x=\dfrac{22\times 3}{19}\]
The above equation can also be written as
\[\Rightarrow \dfrac{1}{10}x=\dfrac{22}{19}\]
On multiplying 10 to both the side of the equation, we get
\[\Rightarrow x=\dfrac{220}{19}\]

Hence, we have solved the equations \[0.3x-0.2y=4\] and \[0.4x+0.5y=\dfrac{63}{19}\] using substitution method and have found the value of x and y. The value of x is \[\dfrac{220}{19}\] and the value of y is \[\dfrac{-50}{19}\].

Note: We should have a better knowledge in the topic of algebra for solving this type of question easily. We should know how to solve the equations using the substitution method. Always remember that if there is a number in which the decimal from the right side is after n digits then, after removing that decimal we will write the number but it will divide the number 10 to the power of n.