Question

How do you solve \[4x-y=5\] and \[x+y=10\] and which method do you use?

Answer 1

Hint: From the question given we have been asked to solve \[4x-y=5\] and \[x+y=10\]. 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.

Complete step by step answer:
From the question given, it has been given that \[4x-y=5\]
Let us assume this equation be \[\left( 1 \right)\].
\[x+y=10\]
Let us assume this equation be \[\left( 2 \right)\].
First of all, let us solve the second equation, \[x+y=10\]
Shift \[x\] from the left hand side of the equation to the right hand side of the equation.
By shifting \[x\] from left hand side of the equation to the right hand side of the equation, we get
\[x+y=10\]
\[\Rightarrow y=10-x\]
Let it be equation \[\left( 3 \right)\]
Now, let us substitute equation \[\left( 3 \right)\] in equation \[\left( 1 \right)\].
By substituting, we get
\[4x-y=5\]
\[\Rightarrow 4x-\left( 10-x \right)=5\]
\[\Rightarrow 4x-10+x=5\]
\[\Rightarrow 5x-10=5\]
Shift \[-10\] from the left hand side of the equation to the right hand side of the equation. By shifting, we get
\[\Rightarrow 5x=5+10\]
\[\Rightarrow 5x=15\]
\[\Rightarrow x=\dfrac{15}{5}\]
\[\Rightarrow x=3\]
Now, substitute \[x=3\] in the equation \[\left( 3 \right)\].
By substituting \[x=3\] in the equation \[\left( 3 \right)\], we get
\[y=10-x\]
\[\Rightarrow y=10-3\]
\[\Rightarrow y=7\]
Therefore, the solution for the given equations is \[x=3\] and\[y=7\].
Hence, by using the substitution method, we got the solution for the given equations.

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. Like this, we have to choose their suitable method to solve the given equations. Calculation should be done very carefully while finding the solution for the given question. Similarly we can solve \[2x-y=3\] and \[x+y=5\] this by substituting $y=2x-3$ in the other equation then we will have $2x-3+x=5\Rightarrow 3x-3=5\Rightarrow x=\dfrac{8}{3}$ and by using this we will have $y=\dfrac{16}{3}-3\Rightarrow y=\dfrac{7}{3}$ .