Solve this system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions.
5x=y=15
3x+2y=9
assuming you mean
5x+y=15
3x+2y=9
multiply the first by 2:
10x + 2y = 30
3x + 2y = 9
now subtract:
7x = 21
x = 3
so, y = 0
solution: (3,0)
To solve this system using elimination, we need to eliminate one variable by manipulating the equations. Let's start by rewriting the system with the correct format:
5x + y = 15 (equation 1)
3x + 2y = 9 (equation 2)
Now, we'll perform elimination to eliminate the variable y. Multiply equation 1 by 2 to make the coefficients of y in both equations equal:
2(5x + y) = 2(15) (equation 1 multiplied by 2)
3x + 2y = 9 (equation 2)
Simplifying the equations, we have:
10x + 2y = 30 (equation 3)
3x + 2y = 9 (equation 2)
Now, subtract equation 2 from equation 3 to eliminate y:
(10x + 2y) - (3x + 2y) = 30 - 9
Simplifying further:
10x - 3x + 2y - 2y = 21x
7x + 0 = 21
7x = 21
Divide both sides by 7:
x = 3
We have found the value of x. Now, substitute x = 3 into one of the original equations (equation 1, for example), to find y:
5x + y = 15
5(3) + y = 15
15 + y = 15
y = 0
Therefore, the solution to the system of equations is (3, 0).