Equations: the "Elimination" Method
Date: 11/20/96 at 19:12:23 From: collin Subject: Easier Ways to Solve Two Equations Solve the system by using the substitution method 5x - 3y = -36 3x - y = -10 This is what I did, but it seems so involved. Is there an easier way? y = -3x - 10 5x - 3(-3x - 10) = -36 5x + 9x + 30 = -36 14x = -66 x = -66/14 x = -33/7 So y = -3(-33/7) -10 y = 29/7
Date: 04/06/97 at 14:25:52 From: Doctor Sydney Subject: Re: Easier Ways to Solve Two Equations Dear Collin, Hello! Thanks for writing. You did a good job solving the problem you wrote about... you have made just one small error, which I will point out below. Sometimes the substitution method CAN be involved, as you note. Sometimes solving two equations with two unknowns or variables can be done in a variety of ways. I must say that the substitution method may not be the best in this particular case. It can certainly be done, but it does seem kind of involved! However, it is good to practice using the substitution method even on problems that seem messy, because it sharpens your skills at doing it! In this case I would use the "elimination" method which you might have already learned about. I'll explain the "elimination" method in detail below. First, though, let's look at your solution to the problem. I have marked where you make the mistake. > ... > y = -3x - 10 ******** OOPS! y = 3x + 10 ******** BUT, ASSUMING THAT Y=-3X - 10, YOU DID A FINE JOB ON THE REST OF IT! So, your work is fine except that you made a sign error. Go through the problem again using the equation y = 3x + 10 to find the right values for x and y. In general, if you want to check your answers to these kinds of problems, plug the values you get for both x and y into both equations to make sure they work. You must make sure that your value works for BOTH equations. If your answer does not work for either of your equations, it is not the right answer. For instance, look at the values you got for x and y above: you got that x = -33/7 and y = 29/7. Now plug those values into your second equation 3x - y = -10. You get 3(-33/7) - (29/7) = -10. But 3(-33/ 7) - (29/7) = -128/7 = 18.29, and 18.29 is not equal to 10, so these values x = -33/7 and y = 29/7 must not be right. Now, as promised, I will explain how to use a method other than substitution to solve this system of equations. This is called the "elimination" method by some people. First, let's write our system of equations down again: 5x -3y = -36 3x -y = -10 Now, our first step is to choose one of the variables (either x or y), and then once we have chosen a variable (say we choose x), we multiply each equation by something so that the coefficients of that variable are the same in both equations. It doesn't matter whether you choose x or y in this first step, although sometimes choosing one over the other will make it easier for you later. After practicing this method, you will get a feel for which variable to choose. For now, let's choose the variable x. Now we want to multiply each of the two equations by a number such that the coefficient of x is the same in each equation. One way to do this is to multiply the first equation by 3 and the second by 5. times 3: 15x - 9y = -108 times 5: 15x - 5y = -50 Now you subtract the second equation from the first. Doing this gives you a new equation with just one variable! Now you can see why we wanted the coefficients of the x variable to be the same... when we subtract equations from one another, the x term goes away. We are left with the following equation: Subtract: -4y = -58 Solve for y to get: y = -58 / -4 = (14 and 1/2) Now that you have a y value, you can substitute this value in for either of your two original equations to get another equation that has only an x variable. Solve this equation to figure out what x is. Again, you should check your answer by making sure your x and y values work for BOTH equations. Solving simultaneous equations is nice in this way, because you check your answer more easily when doing these problems than you can in dealing with other kinds of problems. I hope that this helps. If you need more help, please do write back! -Doctors D. and Sydney, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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