Question

How do you solve the system of equations 3x-4y=9 and 8x+6y=-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-4y=9. Then, add the whole equation by 4y and find the value of 3x. And divide the obtained equation by 3 to find the value of x. Now, consider the equation 8x+6y=-2 and divide the whole equation by 2. Substitute the value of x in the obtained equation. Then, bring all the y terms to the LHS and the constants to the RHS. We get an equation with variable y. multiply the whole equation by 3. Do necessary calculations and find the value of y. Then, substitute the value of y in 3x-4y=9 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-4y=9 and 8x+6y=-2.
We have been given the two equation are
3x-4y=9 ---------(1) and
8x+6y=-2---------(2)
We first have to consider the equation (1).
Add the whole equation by 4y.
\[\Rightarrow 3x-4y+4y=9+4y\]
We know that terms with the same magnitude and opposite signs cancel out. We get
3x=9+4y
Now, divide the whole equation by 3.
 \[\Rightarrow \dfrac{3x}{3}=\dfrac{9+4y}{3}\]
We find that 3 are common in both the numerator and denominator of LHS.
On cancelling 3, we get
\[x=\dfrac{9+4y}{3}\] -----------(3)
Let us now consider equation (2).
Divide the whole equation by 2.
\[\Rightarrow \dfrac{8x+6y}{2}=\dfrac{-2}{2}\]
\[\Rightarrow \dfrac{8x}{2}+\dfrac{6y}{2}=\dfrac{-2}{2}\]
We can write the expression as
\[\dfrac{4\times 2x}{2}+\dfrac{3\times 2y}{2}=\dfrac{-2}{2}\]
We find that 2 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling 2, we get
4x+3y=-1 -----------(4)
Now substitute the value of x in the equation (4).
\[\Rightarrow 4\left( \dfrac{9+4y}{3} \right)+3y=-1\]
Multiply the whole equation by 3. We get
\[\Rightarrow 4\times 3\left( \dfrac{9+4y}{3} \right)+3\times 3y=-1\times 3\]
We find that 3 are common in both the denominator and numerator. Cancelling 3, we get
\[4\left( 9+4y \right)+3\times 3y=-1\times 3\]
\[\Rightarrow 4\left( 9+4y \right)+9y=-3\]
Use distributive property, that is \[a\left( b+c \right)=ab+ac\], we get
\[\Rightarrow 4\times 9+4\times 4y+9y=-3\]
On further simplification, we get
\[\Rightarrow 36+16y+9y=-3\]
On grouping the y terms and adding them, we get
\[36+25y=-3\]
Subtract 36 from both the sides of the equation.
\[\Rightarrow 36+25y-36=-3-36\]
We know that terms with the same magnitude and opposite signs cancel out. We get
25y=-3-36
\[\Rightarrow 25y=-39\]
Divide the whole expression by 25. We get
\[\Rightarrow \dfrac{25y}{25}=\dfrac{-39}{25}\]
We find that 25 are common in both the numerator and denominator of the LHS.
On cancelling 25, we get
\[y=\dfrac{-39}{25}\]
Now, substitute the value of in equation (1).
\[\Rightarrow 3x-4\left( \dfrac{-39}{25} \right)=9\]
On further simplifications we get
\[3x+\dfrac{156}{25}=9\]
Let us now subtract \[\dfrac{156}{25}\] from both the sides of the equation.
\[\Rightarrow 3x+\dfrac{156}{25}-\dfrac{156}{25}=9-\dfrac{156}{25}\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[3x=9-\dfrac{156}{25}\]
Let us take LCM in the RHS. We get
\[\Rightarrow 3x=\dfrac{9\times 25-156}{25}\]
On further simplifications, we get
\[3x=\dfrac{225-156}{25}\]
\[\Rightarrow 3x=\dfrac{69}{25}\]
Let us now divide the expression by 3. We get
\[\dfrac{3x}{3}=\dfrac{69}{3\times 25}\]
We can simplify the equation as
\[\dfrac{3x}{3}=\dfrac{3\times 23}{3\times 25}\]
We find that 3 are common in both the numerator and denominator of both the LHS and RHS.
On cancelling 3, we get
\[x=\dfrac{23}{25}\]

Therefore, the value of x and y in the system of equations 3x-4y=9 and 8x+6y=-2 are \[\dfrac{23}{25}\] and \[\dfrac{-39}{25}\] 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 8x+6y=-2.
Here, LHS=8x+6y
But we know that \[x=\dfrac{23}{25}\] and \[y=\dfrac{-39}{25}\].
Therefore, we get
LHS=\[8\left( \dfrac{23}{25} \right)+6\left( \dfrac{-39}{25} \right)\]
On further simplifications, we get
LHS=\[\dfrac{23\times 8}{25}-\dfrac{39\times 6}{25}\]
LHS=\[\dfrac{184}{25}-\dfrac{234}{25}\]
LHS=\[\dfrac{184-234}{25}\]
LHS=\[\dfrac{-50}{25}\]
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.