Graphing Systems of EquationsDate: 06/01/98 at 16:01:53 From: Kelly Sifferman Subject: How to graph system equations Solve the following system of equations by graphing: 3x + 2y = 5 -3x - 2y = 10 Thanks. Date: 06/01/98 at 16:37:08 From: Doctor Gary Subject: Re: How to graph system equations Do you know how to graph a linear relation between x and y? You can always start with a value for x and then determine the value of y required to satisfy the equation. For example, use the equation: 3x + 2y = 5 If x is 0, then y must be 5/2. If x is 1, then y must be 1. Since two points define a line, all you have to do is draw the line that "connects" (0,5/2) to (1,1). The solution to a system of linear equations is the values of x and y at the point at which the lines intersect. Don't be upset that you can't solve this system, because there is no solution: if you multiply both sides of the second equation by -1, you'll see that there is no solution for this system of equations. 3x + 2y can't be 5 and -10 at the same time. If you graph these two lines, you'll see that they are parallel. You can see this without graphing, by re-expressing each equation in the "standard" form of a linear function in which y is equal to the sum of (slope of the line times x) plus (y co-ordinate of the y-intercept). Your two equations are: 3x + 2y = 5 2y = -3x + 5 y = (-3/2)x + (5/2) and -3x - 2y = 10 -2y = 3x + 10 y = (-3/2)x - 5 The two lines have the same slope, but they have different y-intercepts. -Doctor Gary, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/