Solve the following system of equations by elimination:

\[
3x + 4y = 23
\]

\[
5x + 3y = 31
\]

let 3 x + 4 y = 23 be equation 1 let 5 x + 3 y = 31 be equation 2
do ( e q u 1 ) ∗ 3 and ( e q u 2 ) ∗ 4
now 9 x + 12 y = 69
and 20 x + 12 y = 124 now eliminate the y by doing ( e q u 2 ) − ( e q u 1 ) 11 x = 55 x = 5
now substitute x = 5 back into equation 1 or 2 3 ∗ ( 5 ) + 4 y = 23 4 y = 23 − 15 y = 2

Answered by edmc96 | 2024-06-10

The solution to the system of equations 3x+4y=23 and 5x+3y=31 by elimination is x=5 and y=2 after manipulating the equations to cancel out y and solving for x.
To solve the system of equations 3x+4y=23 and 5x+3y=31 by elimination, we must manipulate the equations to eliminate one of the variables. We'll try to eliminate y. If we multiply the first equation by 3 and the second equation by -4, we'll make the coefficients of y in both equations equal in magnitude but opposite in sign.
First Equation (multiplied by 3): 9x + 12y = 69 Second Equation (multiplied by -4): -20x - 12y = -124
Adding the new versions of the first and second equations gives us: 9x - 20x + 12y - 12y = 69 - 124 -11x = -55
Solving for x, we get: x = 5
Now plug x back into one of the original equations to find y. Using the first equation: 3(5) + 4y = 23 15 + 4y = 23 4y = 8 y = 2
So the solution to the system by elimination method is x=5 and y=2.

Answered by ToddBoyce | 2024-06-24

The solution to the system of equations is x = 5 and y = 2. Using the elimination method, we derived these values by aligning the equations and eliminating one variable at a time. The solution can be verified by substituting these values back into the original equations.
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Answered by edmc96 | 2024-12-26