Question

How do you solve the system of equations 3x+5y=4 and 3x+7y=2?

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+5y=4. Then, subtract the whole equation by 5y and find the value of 3x. Now, consider the equation 3x+7y=2 and subtract the equation by 7y. Equate the values of 3x obtained. Then, bring all the y terms to the LHS and the constants to the RHS. We get an equation with variable y. Add 7y on both the sides of the equation and divide the equation by 2 to find the value of y. Then, substitute the value of y in 3x+5y=4 and obtain the value of x.

Complete step by step solution:
According to the question, we are asked to solve the system of equations 3x+5y=4 and 3x+7y=2.
We have been given the two equation are
3x+5y=4 ---------(1) and
3x+7y=2---------(2)
We first have to consider the equation (1).
Subtract the whole equation by 5y.
\[\Rightarrow 3x+5y-5y=4-5y\]
We know that terms with the same magnitude and opposite signs cancel out. We get
3x=4-5y -----------(3)
Let us now consider equation (2).
Subtract the whole equation by 7y.
\[\Rightarrow 3x+7y-7y=2-7y\]
We know that terms with the same magnitude and opposite signs cancel out. We get
3x=2-7y -----------(4)
From equation (3) and (4), we find that the LHS is the same, that is 3x.
Therefore, we can equate their RHS.
\[\Rightarrow 4-5y=2-7y\]
Now, we have to add 7y on both the sides of the equation. We get
\[4-5y+7y=2-7y+7y\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[\Rightarrow 4-5y+7y=2\]
Now, subtract 4 from both the sides of the equations.
\[\Rightarrow 4-5y+7y-4=2-4\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 4, we get
\[-5y+7y=2-4\]
Now, we have to group the y terms and solve for y.
\[\Rightarrow 2y=2-4\]
On further simplifications, we get
\[2y=-2\]
We have to divide the whole equation by 2.
Therefore, we get
\[\dfrac{2y}{2}=\dfrac{-2}{2}\]
We find that 2 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling 2, we get
y=-1
Now, substitute the value of y in equation (1) to find the value of x.
\[\Rightarrow 3x+5\left( -1 \right)=4\]
On further simplifications, we get
\[3x-5=4\]
Add 5 on both the sides of the equation. We get
3x+5-5=4+5
We know that terms with the same magnitude and opposite signs cancel out. We get
\[\Rightarrow 3x=4+5\]
\[\Rightarrow 3x=9\]
Now, we have to divide the equation by 3.
\[\Rightarrow \dfrac{3x}{3}=\dfrac{9}{3}\]
We can simplify the equation as
\[\Rightarrow \dfrac{3x}{3}=\dfrac{3\times 3}{3}\]
We find that 3 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling 3, we get
\[x=3\]

Therefore, the value of x and y in the system of equations 3x+5y=4 and 3x+7y=2 are 3 and -1 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 (2), that is 3x+7y=2.
Here, LHS=3x+7y
But we know that x=3 and y=-1.
Therefore, we get
LHS=3(3)+7(-1).
On further simplifications, we get
LHS=9-7
Therefore, LHS=2.
Now consider the RHS.
RHS=2
Here, LHS=RHS.
It is enough to check in one linear equation.
Hence the obtained answer is correct.