Question

How do you solve the system of equations: 3x-y=8, y=x-4?

Answer 1

Hint: This type of problem is based on the concept of linear equations with two variables. Here, we first consider the equation 3x-y=8. Then, subtract the whole equation by 3x and cancel -1 from the LHS and RHS. Equate the value of y with the equation y=x-4. Then, bring all the x terms to the LHS and the constants to the RHS. We get an equation with variable x. Subtract x from both the sides of the equation to find the value of x. Then, substitute the value of x in y=x-4 and obtain the value of y.

Complete step by step solution:
According to the question, we are asked to solve the system of equation: 3x-y=8, y=x-4.
We have been given the two equation are
3x-y=8---------(1) and
y=x-4---------(2)
We first have to consider the equation (1).
Subtract the whole equation by 3x.
\[\Rightarrow 3x-y-3x=8-3x\]
We know that terms with same magnitude and opposite signs cancel out. We get
-y=8-3x
Let us now take -1 common from both the sides of the equation.
\[\Rightarrow -1\left( y \right)=-1\left( 3x-8 \right)\]
Divide the whole equation by -1. We get
\[\dfrac{-1\left( y \right)}{-1}=\dfrac{-1\left( 3x-8 \right)}{-1}\]
We find that -1 are common in both the numerator and denominator of both the LHS and RHS.
Let us cancel the common terms.
\[\therefore y=3x-8\] -------------(3)
From equation (2) and (3), we find that the LHS is the same.
Therefore, we can equate their RHS.
\[\Rightarrow x-4=3x-8\]
Now, we have to subtract x from both the sides of the equation. We get
\[x-4-x=3x-8-x\]
We know that terms with same magnitude and opposite signs cancel out. We get
\[\Rightarrow -4=3x-8-x\]
\[\Rightarrow 3x-8-x=-4\]
Now, we have to group the x terms and solve for x.
\[\Rightarrow 2x-8=-4\]
Add 8 on both the sides of the equation. We get
2x-8+8=-4+8
We know that terms with same magnitude and opposite signs cancel out. We get
\[\Rightarrow 2x=-4+8\]
\[\Rightarrow 2x=4\]
Now, we have to divide the equation by 2.
\[\Rightarrow \dfrac{2x}{2}=\dfrac{4}{2}\]
We can simplify the equation as
\[\Rightarrow \dfrac{2x}{2}=\dfrac{2\times 2}{2}\]
We find that 2 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling 2, we get
\[x=2\]
Now, substitute the value of x in y=x-4.
We get,
y=2-4
on further simplification, we get
\[y=-2\]

Therefore, the value of x and y in the system of equations 3x-y=8 and y=x-4 are 2 and -2 respectively.

Note: We can verify whether the answer obtained is correct or not.
Substitute the values of x and y in the given system and check whether the LHS is equal to RHS.
Consider equation (1), that is 3x-y=8.
Here, LHS=3x-y
But we know that x=2 and y=-2.
Therefore, we get
LHS=3(2)-(-2).
On further simplifications, we get
LHS=6+2
Therefore, LHS=8.
Now consider the RHS.
RHS=8
Here, LHS=RHS.
It is enough to check in one linear equation.
Hence the obtained answer is correct.