I cannot figure this problem out
1 answer:
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Answer:
(x, y) = (3, -6)
Step-by-step explanation:
I like a good graphing calculator for solving systems of equations by graphing.
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If you're solving these by hand, you need to graph the equations. It can be convenient to put the equations into "intercept form" so you can use the x- and y-intercepts to draw your graph.
That form is ...
x/(x-intercept) +y/(y-intercept) = 1
Dividing a standard-form equation by the constant on the right will put it in this form.
x/(-12/2) +y/(-12/3) = 1 . . . . . . divide the first equation by -12
x/(-6) +y/(-4) = 1 . . . . . . . . . . . the x-intercept is -6; the y-intercept is -4
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x/(12/10) +y/(12/3) = 1 . . . . . . divide the second equation by 12
x/1.2 +y/4 = 1 . . . . . . . . . . . . . the x-intercept is 1.2*; the y-intercept is 4
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The locations of these intercepts and the slopes of the lines tell you that the solution will be in the fourth quadrant. The lines intersect at (x, y) = (3, -6).
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* It can be difficult to draw an accurate graph using an intercept point that is not on a grid line. It may be desirable to put the second equation into slope-intercept form, so you can see the rise/run values that let you choose grid points on the line. That equation is y =-10/3x +4. A "rise" of -10 for a "run" of +3 will get you to (3, -6) starting from the y-intercept of (0, 4).
so the 2nd equation is really the first equation in disguise.
since both equations are the same, that means if you graph them, is just one line pancaked over the other, and the solutions points is every single one on each, namely infinitely many solutions.