y = x^2 + 4x + 1
y = 2x + 1
solve each system of equations algebraically?
2009-03-24 11:11 am
回答 (4)
2009-03-24 11:31 am
✔ 最佳答案
y = x^2 + 4x + 1 equation1y = 2x + 1 equation 2
plug in equation 2 to equation 1
2x+1=x^2+4x+1
x^2+4x-2x+1-1=0
x^2+2x=0
x(x+2)=0
x=0 x=-2 answer
x=0
y=2(0)+1
y=1 answer//
x=-2
y=2(-2)+1
y=-4+1
y=-3 answer//
cheking
1=(0)^2+4(0)+1
1=1 correct!
2015-08-13 8:45 pm
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RE:
solve each system of equations algebraically?
y = x^2 + 4x + 1
y = 2x + 1
RE:
solve each system of equations algebraically?
y = x^2 + 4x + 1
y = 2x + 1
參考: solve system equations algebraically: https://biturl.im/oFNZy
2009-03-24 12:05 pm
y = x^2 + 4x + 1 (solve by using substitution)
y = 2x + 1
y = x^2 + 4x + 1
2x + 1 = x^2 + 4x + 1
x^2 + 4x - 2x - 1 + 1 = 0
x^2 + 2x = 0
x(x + 2) = 0
x = 0
x + 2 = 0
x = -2
y = 2x + 1
y = 2(-2) + 1
y = -4 + 1
y = -3
y = 2x + 1
y = 2(0) + 1
y = 0 + 1
y = 1
∴ (x = -2 , y = -3) , (x = 0 , y = 1)
y = 2x + 1
y = x^2 + 4x + 1
2x + 1 = x^2 + 4x + 1
x^2 + 4x - 2x - 1 + 1 = 0
x^2 + 2x = 0
x(x + 2) = 0
x = 0
x + 2 = 0
x = -2
y = 2x + 1
y = 2(-2) + 1
y = -4 + 1
y = -3
y = 2x + 1
y = 2(0) + 1
y = 0 + 1
y = 1
∴ (x = -2 , y = -3) , (x = 0 , y = 1)
2009-03-24 11:27 am
First i would like to know, is it in your school home work? then u should try it in ur own. anyway i have given the answer but from the next time try urself. good luck
y = x^2 + 4x +1
or 2x + 1 = x^2 + 4x + 1 [as y = 2x + 1]
or x^2 + 4x - 2x + 1 - 1 = 0
or x^2 + 2x = 0
or x(x + 2) = 0
either x = 0
therefore y = 2x + 1 = 1
or x + 2 = 0
or x = -2
therefore y = 2x + 1 = 2(-2) + 1 = -4 + 1 = -3
y = x^2 + 4x +1
or 2x + 1 = x^2 + 4x + 1 [as y = 2x + 1]
or x^2 + 4x - 2x + 1 - 1 = 0
or x^2 + 2x = 0
or x(x + 2) = 0
either x = 0
therefore y = 2x + 1 = 1
or x + 2 = 0
or x = -2
therefore y = 2x + 1 = 2(-2) + 1 = -4 + 1 = -3
2009-03-24 11:25 am
it is a silmultaneous equation, use substitution method
so 2x + 1 = x^2 + 4x + 1
thus 0 = x^2 + 4x + 1 - 2x - 1
thus 0 = x^2 + 2x
thus 0 = x(x+2)
thus x = 0 or x = -2
if x = 0 y = 1
and if x = 1 y = 3
so 2x + 1 = x^2 + 4x + 1
thus 0 = x^2 + 4x + 1 - 2x - 1
thus 0 = x^2 + 2x
thus 0 = x(x+2)
thus x = 0 or x = -2
if x = 0 y = 1
and if x = 1 y = 3
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