Solving Linear Systems Algebraically KoreyAnne Smith.

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Presentation transcript:

Solving Linear Systems Algebraically KoreyAnne Smith

1. Solve equation for x 2. Substitute the expression for x into the equation and solve for y 3. Substitute the value y into the new equation and solve for x 4. CHECK FOR MISTAKES!! KoreyAnne Smith

PPROBLEM: 2x – 4y = 18 3x – y = 22 SSET THE SECOND EQUATION EQUAL TO Y:  -y – 3x + 22  y = 3x – 22

SSUBSTITUTE THE Y INTO THE EQUATION: 2x – 4(3x-22) = 18 BBEGIN SOLVINGTHE PROBLEM: 2x -12x + 88 = x + 88 = 18 x = 7

 SOLVING FOR Y:  PLUG X INTO THE EQUATION: y = 3(7) – 22 y = 21 – 22 y = -1  ANSWER: (7,-1) KoreyAnne Smith

1. Multiply the first equation by the coefficient so that they only differ in sign 2. Add the revised equations and solve for y 3. Substitute the value of y into one of the original equations 4. Solve for x 5. CHECK FOR MISTAKES!!! KoreyAnne Smith

 PROBLEM: x + 2y = -3 -3x + y = -19  MAKE THE COEFFICIENTS OPPOSITE:  3(x + 2y = -3)  -3x + y = -19 KoreyAnne Smith

 NEW PROBLEM: 3x + 6y = x + y = -19 7y = -28  SOLVE FOR Y: 7y = y = -4 KoreyAnne Smith

 SOLVING FOR X: o Go back to an original equation and plug in y: x + 2y = -3  NEW EQUATION: x + 2(-4) = -3 x -8 = -3 x = 5  ANSWER: (5, -4) KoreyAnne Smith