Question 269165
Here's a detail steps for solving the equation! ☺
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5x-3y=24_3x+5y=28

Multiply each equation by the value that makes the coefficients of y equal.  This value is found by dividing the least common multiple of the coefficients of y by the current coefficient.  In this case, the least common multiple is 15.
5*(5x-3y=24)_3*(3x+5y=28)

Multiply each equation by the value that makes the coefficients of y equal.  This value is found by dividing the least common multiple of the coefficients of y by the current coefficient.  In this case, the least common multiple is 15.
5*(5x-3y)=5(24)_3*(3x+5y)=3(28)

Multiply 5 by each term inside the parentheses.
5*(5x-3y)=120_3*(3x+5y)=3(28)

Multiply 5 by each term inside the parentheses.
(25x-15y)=120_3*(3x+5y)=3(28)

Remove the parentheses around the expression 25x-15y.
25x-15y=120_3*(3x+5y)=3(28)

Multiply 3 by each term inside the parentheses.
25x-15y=120_3*(3x+5y)=84

Multiply 3 by each term inside the parentheses.
25x-15y=120_(9x+15y)=84

Remove the parentheses around the expression 9x+15y.
25x-15y=120_9x+15y=84

Add the two equations together to eliminate y from the system.
 9x+15y=84_<U>25x-15y=120<u>_34x    =204

Divide each term in the equation by 34.
x=6

Substitute the value found for x into the original equation to solve for y.
25(6)-15y=120

Multiply 25 by each term inside the parentheses.
150-15y=120

Move all terms not containing y to the right-hand side of the equation.
-15y=-30

Divide each term in the equation by -15.
y=2

This is the final solution to the independent system of equations.
Answer: x=6
Answer: y=2