Final answer:
The system of equations 5x+3y=12 and x-4y=7 can be solved using the elimination method. Multiply one or both equations to make the coefficients opposite for one variable, add or subtract the new equations to eliminate that variable, and solve for the remaining variable. The solutions are x = 20/7 and y = 44/7.
Step-by-step explanation:
To solve the system of equations 5x+3y=12 and x-4y=7 using elimination, we want to eliminate one variable by multiplying one or both equations by appropriate constants so that the coefficients of one variable are equal in magnitude but opposite in sign. In this case, we can multiply the second equation by 3 to make the coefficients of y opposite: 3(x-4y) = 3(7). This gives us 3x - 12y = 21. Now, we can add this new equation to the first equation to eliminate y. The result is 5x + 3x - 12y = 12 + 21, which simplifies to 8x - 12y = 33.
Next, we can divide this equation by 4 to make the coefficient of x in front of one variable 1: 2x - 3y = 8. Now we have a system of equations: 2x - 3y = 8 and 5x + 3y = 12. To eliminate y, we can add these two equations together. This gives us (2x-3y) + (5x+3y) = 8 + 12, which simplifies to 7x = 20.
Finally, we can solve for x by dividing both sides of the equation by 7: x = 20/7. Substitute this value back into one of the original equations to solve for y. Using the first equation, we have: 5(20/7) + 3y = 12. Simplifying, we get 100/7 + 3y = 12. Subtracting 100/7 from both sides and simplifying further, we find that y = 12 - 100/7 = 44/7.