Solve the system of equations using the linear combination method (addition, not subtraction). Show ALL steps used to reach your answer. {4x + 3y = 7 {2x + y = 5 Solve the system of equations using the linear combination method (addition, not subtraction). Show ALL steps used to reach your answer. {4x + 3y = 7 {2x + y = 5
The Correct Answer and Explanation is :
To solve the system of equations using the linear combination method (also known as the addition method), we aim to eliminate one of the variables by adding or subtracting the equations. We proceed step-by-step:
Given system of equations:
- ( 4x + 3y = 7 )
- ( 2x + y = 5 )
Step 1: Multiply the second equation to align the coefficients
Since the coefficient of ( y ) in the second equation is ( 1 ), and the coefficient of ( y ) in the first equation is ( 3 ), we can eliminate ( y ) by multiplying the second equation by ( 3 ) to match the ( y )-terms in both equations.
Multiply the entire second equation by ( 3 ):
[
3(2x + y) = 3(5)
]
This gives:
[
6x + 3y = 15
]
Now, we have the modified system of equations:
- ( 4x + 3y = 7 )
- ( 6x + 3y = 15 )
Step 2: Subtract the first equation from the second
Now that both equations have the same coefficient for ( y ), we subtract the first equation from the second equation to eliminate ( y ).
[
(6x + 3y) – (4x + 3y) = 15 – 7
]
Simplifying:
[
6x – 4x + 3y – 3y = 8
]
[
2x = 8
]
Step 3: Solve for ( x )
Now, solve for ( x ):
[
2x = 8
]
[
x = \frac{8}{2}
]
[
x = 4
]
Step 4: Substitute ( x = 4 ) into one of the original equations
Now that we know ( x = 4 ), substitute this value into either of the original equations to solve for ( y ). Let’s use the second equation:
[
2x + y = 5
]
Substitute ( x = 4 ):
[
2(4) + y = 5
]
[
8 + y = 5
]
[
y = 5 – 8
]
[
y = -3
]
Step 5: Final solution
Thus, the solution to the system of equations is:
[
x = 4, \quad y = -3
]
Explanation:
The linear combination method involves manipulating the equations to make one of the variables cancel out when the equations are added or subtracted. In this case, by multiplying the second equation by 3, we created a system where the coefficients of ( y ) were the same, allowing us to eliminate ( y ) by subtraction. Once we solved for ( x ), we substituted the value of ( x ) back into one of the original equations to find ( y ). This method is effective for solving systems where one of the variables can be easily eliminated by addition or subtraction.