Question

How do you solve the system of equations 4x+2y=16 and 3x+3y=18?

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+3y=18. Then, divide the whole equation by 3 and subtract x from the LHS and RHS. Substitute the value of y in the equation 4x+2y=16. Then, bring all the x terms to the LHS and the constants to the RHS. We get an equation with variable x. Add 2x on both sides of the equation to find the value of x. Then, substitute the value of x in 3x+3y=18 and obtain the value of y.

Complete step by step solution:
According to the question, we are asked to solve the system of equations: 4x+2y=16 and 3x+3y=18.
 We have been given the two equation are
4x+2y=16 ---------(1) and
3x+3y=18---------(2)
We first have to consider the equation (2).
Divide the whole equation by 3.
\[\Rightarrow \dfrac{3x+3y}{3}=\dfrac{18}{3}\]
\[\Rightarrow \dfrac{3x}{3}+\dfrac{3y}{3}=\dfrac{18}{3}\]
We find that 3 are common in both the numerator and denominator of the LHS. On cancelling the common term, we get
\[x+y=\dfrac{18}{3}\]
We can simplify the RHS as
\[x+y=\dfrac{6\times 3}{3}\].
Let us cancel the common term 3 from the numerator and denominator.
\[\Rightarrow x+y=6\]
Now, subtract x from both the sides of the equation.
\[\Rightarrow x+y-x=6-x\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[y=6-x\] ----------(3)
Now, we have to consider equation (1).
Divide equation (1) by 2 on both the sides.
\[\Rightarrow \dfrac{4x+2y}{2}=\dfrac{16}{2}\]
\[\Rightarrow \dfrac{4x}{2}+\dfrac{2y}{2}=\dfrac{16}{2}\]
We can further simplify the equation as
\[\dfrac{2\times 2x}{2}+\dfrac{2y}{2}=\dfrac{8\times 2}{2}\]
We find that 2 are common in both the numerator and denominator of both the LHS and RHS.
Let us cancel the common terms.
\[\Rightarrow 2x+y=8\]
Now, we have to subtract 2x from both the sides of the equation. We get
\[2x+y-2x=8-2x\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[\Rightarrow y=8-2x\] ----------(4)
From equation (3) and (4), we find that the LHS is the same.
Therefore, we can equate their RHS.
\[\Rightarrow 6-x=8-2x\]
Now, we have to add 2x on both sides of the equation. We get
\[6-x+2x=8-2x+2x\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[\Rightarrow 6-x+2x=8\]
\[\Rightarrow 6+x=8\]
Now let us subtract 6 from both the sides of the equation. We get
\[\Rightarrow 6+x-6=8-6\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[\Rightarrow x=8-6\]
\[\therefore x=2\]
Now, substitute the value of x in 3x+3y=18.
We get,
\[3\times 2+3y=18\]
\[\Rightarrow 6+3y=18\]
Subtract 6 from both the sides of the equation. We get
\[\Rightarrow 6+3y-6=18-6\]
We know that terms with the same magnitude and opposite signs cancel out. We get
\[3y=18-6\]
\[\Rightarrow 3y=12\]
Divide the whole equation by 3. We get
\[\Rightarrow \dfrac{3y}{3}=\dfrac{12}{3}\]
We can further simplify the equation as
\[\dfrac{3y}{3}=\dfrac{4\times 3}{3}\]
We find that 3 are common in both the numerator and denominator of both the LHS and the RHS.
Cancelling the common term 3, we get
\[y=4\]

Therefore, the value of x and y in the system of equations 4x+2y=16 and 3x+3y=18 are 2 and 4 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 4x+2y=16.
Here, LHS=4x+2y
But we know that x=2 and y=4.
Therefore, we get
LHS=4(2)+2(4).
On further simplifications, we get
LHS=8+8
Therefore, LHS=16
Now consider the RHS.
RHS=16
Here, LHS=RHS.
It is enough to check in one linear equation.
Hence the obtained answer is correct.