By now you should see that our answer for x will be ZERO.ĭo you see what I see? Yes, I found the SAME point of intersection again, which is (0,10). Next: To find x, we plug y = 10 into EITHER of the original equations. See it? See why? Here's why: A negative PLUS a positive = ZERO. In our second 3x + 2y = 20, you can eliminate 3x by multiplying -3 by EVERY term in our first equation (x + y = 10). GOAL: Eliminate x and solve for y or vice-versa. Keep in mind that it is your choice which variable you want to eliminate first. This method deals with matching the variables to ELIMINATE or do away with one. So, our point of intersection is once again (0,10). Our second equation was \(3x + 2y = 20\) and, after substituting, becomes \(3x + 2(-x + 10 ) = 20\)ģ) Plug x = 0 into EITHER original equations to find the value of y. So, \(x + y = 10\) becomes \(y = -x + 10\).Ģ) Plug the value of y (that is, -x + 10) in the second equation to find x. Here's what these two equations look like on the xy-plane: Method 2: Solve algebraicallyġ) Solve for eaither x or y in the first equation (\(x + y = 10\)). Point (0,10) means that if you plug x = 0 and y = 10 into BOTH original equations, you will find that it solves both equations. After graphing these lines, you'll find that BOTH equations meet at point (0,10). Then, graph the two lines, leading to the point of intersection. Next, \(3x + 2y = 20\) becomes \(y = -\frac + 10\) when written in slope-intercept form. To solve graphically, it is best to write BOTH equations in the slope-intercept form or in the form: \(y = mx + b\) where m = the slope and b = the y-intercept as your first step. I will solve the question using all 3 methods.
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