Question

How do you solve the system of equations \[3x+7y=1\] and \[–x+y=-7\]?

Answer 1

Hint: This type of problem is based on the concept of linear equations with two variables. Here, we first consider the equation \[–x+y=-7.\] Then, add the whole equation by x and find the value of y. Now, substitute the value of y in the equation \[3x+7y=1.\] Then, bring all the x terms to the LHS and the constants to the RHS. We get an equation with variable x. Find the value of x with necessary calculations. Then, substitute the value of x in \[–X+Y=-7\] and obtain the value of y.

Complete step by step solution:
According to the question, we are asked to solve the system of equations \[3x+7y=1\] and \[–x+y=-7\].
We have been given the two equation are
3x+7y=1 ---------(1) and
–x+y=-7---------(2)
We first have to consider the equation (2).
Add the whole equation by x.
\[\Rightarrow -x+y+x=-7+x\]
We know that terms with the same magnitude and opposite signs cancel out. We get
y=-7+x
We can write the equation as
y=x-7 --------------(3)
Now, substitute the value of y from equation (3) to equation (1).
Therefore, we get
\[3x+7\left( x-7 \right)=1\]
Now, we have to use the distributive property, that is \[a\left( b+c \right)=ab+ac\], to solve the equation further.
\[\Rightarrow 3x+7x-7\times 7=1\]
We know that \[7\times 7=49\].
 Substituting this value in the above equation, we get
\[3x+7x-49=1\]
On grouping the x terms and adding them, we get
\[10x-49=1\]
Now, we have to add 49 on both the sides of the equation.
\[\Rightarrow 10x-49+49=1+49\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[\Rightarrow 10x=1+49\]
\[\Rightarrow 10x=50\]
Now, we have to divide the whole equation by 10. We get,
\[\dfrac{10x}{10}=\dfrac{50}{10}\]
We can write the equation as
\[\dfrac{10x}{10}=\dfrac{5\times 10}{10}\]
We find that 10 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling 10, we get
x=5
Now, substitute the value of x in –x+y=-7.
We get,
-5+y=-7
Let us add 5 on both the sides of the equation.
\[\Rightarrow -5+y+5=-7+5\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[y=-7+5\]
On further simplifications, we get
\[y=-2\]

Therefore, the value of x and y in the system of equations 3x+7y=1 and –x+y=-7 are 5 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+7y=1.
Here, LHS=3x+7y
But we know that x=5 and y=-2.
Therefore, we get
LHS=3(5)+7(- 2).
On further simplifications, we get
LHS=15-14
Therefore, LHS=1.
Now consider the RHS.
RHS=1
Here, LHS=RHS.
It is enough to check in one linear equation.
Hence the obtained answer is correct.