set the equations equal (both equal y)
3x+20=x^2+2x
x^2-x-20=0
(x-5)(x+4)=0
solutions at x=5, and x=-4
Now put those solutions into either equation, and you have a x,y set.
y=x^2+2x
(-4,8) and (5,35) are the two solutions.
24. Solve the system of equations algebraically. Show all of your steps.
Y = x^2 + 2x and y = 3x + 20. Btw this is a workpad problem. I just need to be put in the right direction, on how to do it.
5 answers
Aww. Thanks Guys! I really appreiciate it.
Wait how did bobpusrsly get x^2-x-20=0
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To get x^2-x-20=0, you can set the two given equations equal to each other because they are both equal to y.
So, y = x^2 + 2x becomes:
y = 3x + 20 (equation 1)
x^2 + 2x = 3x + 20 (substituting y with 3x + 20 from equation 1)
x^2 - x - 20 = 0 (subtracting 3x and 20 from both sides)
This is a quadratic equation in standard form (ax^2 + bx + c = 0) where a=1, b=-1, and c=-20. To solve for x, you can factor or use the quadratic formula.
So, y = x^2 + 2x becomes:
y = 3x + 20 (equation 1)
x^2 + 2x = 3x + 20 (substituting y with 3x + 20 from equation 1)
x^2 - x - 20 = 0 (subtracting 3x and 20 from both sides)
This is a quadratic equation in standard form (ax^2 + bx + c = 0) where a=1, b=-1, and c=-20. To solve for x, you can factor or use the quadratic formula.