Math  /  Algebra

QuestionSolve the system using addition/elimination method. If there is exactly one solution, write as an ordered pair. If not, choose one of the other options. {2x2y=123x+5y=20\left\{\begin{array}{l} 2 x-2 y=-12 \\ -3 x+5 y=20 \end{array}\right. One solution: \square No solution Infinite number of solutions

Studdy Solution

STEP 1

1. We are given a system of two linear equations with two variables.
2. The goal is to find the values of x x and y y that satisfy both equations simultaneously.
3. The addition/elimination method involves adding or subtracting equations to eliminate one of the variables.

STEP 2

1. Align the equations for elimination.
2. Eliminate one variable.
3. Solve for the remaining variable.
4. Substitute back to find the other variable.
5. Check for consistency and determine the type of solution.

STEP 3

Align the given system of equations:
2x2y=123x+5y=20\begin{array}{c} 2x - 2y = -12 \\ -3x + 5y = 20 \end{array}

STEP 4

To eliminate one of the variables, we can multiply the first equation by 3 and the second equation by 2, so that the coefficients of x x will be opposites:
Multiply the first equation by 3:
3(2x2y)=3(12) 3(2x - 2y) = 3(-12) 6x6y=36 6x - 6y = -36
Multiply the second equation by 2:
2(3x+5y)=2(20) 2(-3x + 5y) = 2(20) 6x+10y=40 -6x + 10y = 40

STEP 5

Add the two equations to eliminate x x :
6x6y=36+(6x+10y=40)0x+4y=4\begin{array}{c} 6x - 6y = -36 \\ + (-6x + 10y = 40) \\ \hline 0x + 4y = 4 \end{array}
This simplifies to:
4y=4 4y = 4

STEP 6

Solve for y y :
Divide both sides by 4:
y=44 y = \frac{4}{4} y=1 y = 1

STEP 7

Substitute y=1 y = 1 back into one of the original equations to solve for x x . We'll use the first equation:
2x2(1)=12 2x - 2(1) = -12 2x2=12 2x - 2 = -12
Add 2 to both sides:
2x=10 2x = -10
Divide by 2:
x=5 x = -5

STEP 8

Check the solution by substituting x=5 x = -5 and y=1 y = 1 into both original equations to ensure they are satisfied:
First equation:
2(5)2(1)=12 2(-5) - 2(1) = -12 102=12 -10 - 2 = -12 12=12 -12 = -12 (True)
Second equation:
3(5)+5(1)=20 -3(-5) + 5(1) = 20 15+5=20 15 + 5 = 20 20=20 20 = 20 (True)
Both equations are satisfied, so the solution is consistent.
The solution is:
(5,1) \boxed{(-5, 1)}

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