solve system: (x+2)^2+(y-1)^2=16 and 2x+4y=8
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What methods have you been taught to solve simultaneous equations - substitution method - elimination method - something else ?
What are your thoughts?
Please share your work with us ...even if you know it is wrong.
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Well my first attempt was to just set the bottom problem x and y equal to zero and I got -2 for x and 4 for y but those didn't work for the top problem. This problem is weird because most of this chapter has been about matrixes but this one on the review for our test and I'm not sureña how to do it. I also found that x equals 2 and y equals 1 works for both problems but not sure of the work to get there.
JamesMen45
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Hope this helps
I solved the equation on my own. Im pretty sure its right but dont take my word for it.
I solved the equation on my own. Im pretty sure its right but dont take my word for it.
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These equations do not represent two "problems", a "bottom" and a "top" one; they represent one exercise, consisting of one system of two (non-linear) equations. To learn what this sort of system of equations is and how to solve it, try here.
By the way, plugging zero in for each of the variables and solving for the concurrent value of the other variable is how they taught you, back in beginning algebra, to find the x- and y-intercepts. (here) It has nothing to do with solving systems of equations. Instead, try using substitution techniques, similar to what they taught you for solving systems of linear equations. (here)
There is no reason to expect the x- and y-intercepts of the linear equation to be the x- or y-intercepts of the non-linear equation, nor any reason to expect the x- or y-intercept to be a solution to a system.
Tests are not restricted to covering only and exactly the most recent topics encountered in class. Tests may, and often do, review past material. If you have completely forgotten how to solve systems of equations (and it sounds as though you have), please study at least the lessons at the links. Clearly your instructor thinks that you've seen this material, and expects you to know it, so you can rest assured that you'll be seeing this again, including on the final!
pythagorean314159
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what are we dealing with?
I would recommend graphing both equations and seeing what points the graphs intersect at. Notice that in the first equation, you have a circle, and in the second equation, you have a line. A circle on a coordinate grid has an equation of form (x - h)2 + (y - k)2 = r2, in which (h, k) is the center and r is the radius. That should help you graph the first equation.
I would recommend graphing both equations and seeing what points the graphs intersect at. Notice that in the first equation, you have a circle, and in the second equation, you have a line. A circle on a coordinate grid has an equation of form (x - h)2 + (y - k)2 = r2, in which (h, k) is the center and r is the radius. That should help you graph the first equation.
HallsofIvy
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As Denis said, from 2x+ 4= 8, you get y= (4- x)/2 or, equivalently, y= 2- x/2. Put that in to the second equation:
\(\displaystyle (x+2)^2+(y-1)^2=16\) to get \(\displaystyle (x+ 2)^2+ (2- x/2- 1)^2= (x+ 2)^2+ (1- x/2)^2= 16\).
Multiply out those two squares and combine "like terms" to get a quadratic equation for x.
\(\displaystyle (x+2)^2+(y-1)^2=16\) to get \(\displaystyle (x+ 2)^2+ (2- x/2- 1)^2= (x+ 2)^2+ (1- x/2)^2= 16\).
Multiply out those two squares and combine "like terms" to get a quadratic equation for x.