Substituting in Linear EquationsDate: 11/14/2001 at 19:22:10 From: Jeannette Subject: Substituting in Linear Equations Okay, I am given a question where I have two equations 2y = 5x+8x 3y = 4x+9x Now I am told to substitute "something" but I don't understand anything about solving linear equations. If you could give me the easiest way to understand this stuff, I would greatly appreciate it. Date: 11/17/2001 at 23:32:48 From: Doctor Jeremiah Subject: Re: Substituting in Linear Equations Hi Jeannette, Sets of equations are EASY to solve, once you know the secret. There are only four steps to do: Lets say the equations that you wanted to solve were: 2y = 5x + 8 4y = 3x + 9 Step 1: Solve any equation for any unknown value ------------------------------------------------ It doesn't matter which one. Just pick the one that looks the easiest. Obviously if an equation already has an unknown value isolated on one side of the equals sign, then you are already done with this step. So, for example, if one of your equations was 2y = 5x + 8 you could do this: 2y = 5x + 8 ( 2y )/2 = ( 5x + 8 )/2 2y/2 = 5x/2 + 8/2 y = 5x/2 + 4 Or if one of your equations was x = 3y + 10 then you are already done with this step because it is already solved for x. Lets say that our equation was 2y = 5x + 8 and we solved it to be y = 5x/2 + 4. Step 2: Substitute Step 1 into the other equation ------------------------------------------------- Using that answer from Step 1, take the other equation and replace all occurrences of the unknown value you solved for in Step 1. So, for example, if you finished Step 1 with y = 5x/2 + 4 and your other equation was 4y = 3x + 9, then you would replace all values of y in 4y = 3x + 9 with what you found y was equal to (5x/2 + 4) in step 1. Basically you would do: 4y = 3x + 9 <== y = 5x/2 + 4 4(5x/2 + 4) = 3x + 9 See how we substituted the answer from Step 1 into the other equation? Now we can simplify that: 4y = 3x + 9 <== y = 5x/2 + 4 4(5x/2 + 4) = 3x + 9 20x/2 + 16 = 3x + 9 10x + 16 = 3x + 9 10x + 16 - 3x = 3x + 9 - 3x 7x + 16 = 9 7x + 16 - 16 = 9 - 16 7x = -7 x = -1 Step 3: Substitute Step 2 into Step 1 ------------------------------------- Using that answer from Step 2, take the answer from Step 1 and replace all occurrences of the unknown value you solved for in Step 2. So, for example, if you finished Step 2 with x = -1 and your answer for Step 1 was y = 5x/2 + 4, then you would replace all values of x in y = 5x/2 + 4 with what you found x was equal to (-1) in step 2. Basically you would do: y = 5x/2 + 4 <== x = -1 y = 5(-1)/2 + 4 y = -5/2 + 4 y = -5/2 + 8/2 y = 3/2 Step 4: Check your answer ------------------------- We think x = -1 and y = 3/2 but we need to check them by substituting both of them into both of the original equations and seeing if they are true: 2y = 5x + 8 <== x = -1 y = 3/2 2(3/2) = 5(-1) + 8 3 = -5 + 8 3 = 3 4y = 3x + 9 <== x = -1 y = 3/2 4(3/2) = 3(-1) + 9 6 = -3 + 9 6 = 6 They are both true, so our answers of x = -1 and y = 3/2 must be right. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 08/20/2003 at 21:26:15 From: Johnny Ruiz Subject: Jeanette's Original Equation Dr. Math, I'm confused! Jeanette's problem was 2y = 5x+8x and 3y = 4x+9x. How can 2y = 13x AND 3y ALSO equal 13x? It seems to me that if 2y =13x then 3y must equal more than 13x. Where am I going wrong? If I plug in 4, say, for x, then: 5x=20 and 8x=32 and 2y must equal 52. Then y=26. Then, if 4x=16 and 9x=36, 3y must equal 52. In this case y=17 and 1/3. BUT WHICH IS TRUE? Date: 08/21/2003 at 09:02:34 From: Doctor Peterson Subject: Re: Jeanette's Original Equation Hi, Johnny. Dr. Jeremiah showed how to solve systems of equations in general, but didn't deal with the special features of this system (perhaps suspecting that it had been mistyped, because it was written in an odd way, lacking constant terms). Let's look at the original problem. If we follow the substitution method he showed, we do this: 2y = 5x+8x 3y = 4x+9x First simplify the equations: 2y = 13x 3y = 13x Solve the first equation for y: y = 13/2 x Plug that in the second equation: 3(13/2 x) = 13x 39/2 x = 13x Simplify that, by subtracting 13x from both sides: 13/2 x = 0 Multiply both sides by 2/13: x = 0 Plug that in the equation for y: y = 0 And that's the solution. As you suggested, 2y and 3y will normally be different - if y is positive, 3y will be greater, and if y is negative, 2y will be greater (less negative). But if y is zero, then they will in fact be equal. So your method of solving would be to set to expressions for 13x equal: 2y = 3y 0 = y So you see that the normal method of solving does work here; your mistake was in letting yourself get confused by an unstated assumption that x and y are positive. One of the benefits of algebra is that it lets you solve a problem without having to think about what it means, and getting confused by your expectations. Then when you see the answer, you can look back and think about what happened. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/