Two Equations with Two UnknownsDate: 05/24/99 at 01:00:47 From: Margie Yeager Subject: Algebra 2 Solve by adding or subtracting: 1) 2x-3y = 50 7x+8y = -10 2) 5x+6y = 16 3x-4y = 2 3) 3x+2y = 14 2x+3y = 1 4) 2x+3y = 10 5x-4y = 2 5) 10x+3y = 1 3x-2y = -10 6) 5x+2y = 9 2x+3y = -3 Date: 05/24/99 at 18:03:00 From: Doctor Rick Subject: Re: Algebra 2 Hi, Margie. I will do the first of your problems to show you the technique. Before you add or subtract, you want to multiply one or both equations by constants (that is, multiply both sides of one equation by the same number). Your goal is to eliminate one variable when you do the addition or subtraction. Let's eliminate the y. To do this, we want to have the coefficient of y in the two equations be either the same or the opposite. One way to do this is to multiply each equation by the coefficient of y in the OTHER equation. Multiply the first equation on both sides by 8: 16x - 24y = 400 Multiply the second equation on both sides by 3: 21x + 24y = -30 Now the coefficients of y are opposite, so we can add the two equations: 16x - 24y = 400 21x + 24y = -30 --------------- 37x + 0y = 370 The y term has vanished: 37x = 370 Divide both sides by 37: x = 370/37 = 10 We aren't done yet because we only have a value for x, not for y. Choose either of the original equations, substitute x = 10, then solve for y: 2x - 3y = 50 2(10) - 3y = 50 20 - 3y = 50 -3y = 50 - 20 = 30 y = 30/(-3) = -10 The solution set is { x = 10, y = -10 }. Last but not least, check the answer by substituting these values in both equations: 2x - 3y = 50 7x + 8y = -10 2(10) - 3(-10) = 50 7(10) + 8(-10) = -10 20 + 30 = 50 70 - 80 = -10 50 = 50 -10 = -10 We're done! Now it's your turn. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/