Answer:
Second choice:
[tex]x=2t[/tex]
[tex]y=4t^2+4t-3[/tex]
Fifth choice:
[tex]x=t+1[/tex]
[tex]y=t^2+4t[/tex]
Step-by-step explanation:
Let's look at choice 1.
[tex]x=t+1[/tex]
[tex]y=t^2+2t[/tex]
I'm going to subtract 1 on both sides for the first equation giving me [tex]x-1=t[/tex]. I will replace the [tex]t[/tex] in the second equation with this substitution from equation 1.
[tex]y=(x-1)^2+2(x-1)[/tex]
Expand using the distributive property and the identity [tex](u+v)^2=u^2+2uv+v^2[/tex]:
[tex]y=(x^2-2x+1)+(2x-2)[/tex]
[tex]y=x^2+(-2x+2x)+(1-2)[/tex]
[tex]y=x^2+0+-1[/tex]
[tex]y=x^2[/tex]
So this not the desired result.
Let's look at choice 2.
[tex]x=2t[/tex]
[tex]y=4t^2+4t-3[/tex]
Solve the first equation for [tex]t[/tex] by dividing both sides by 2:
[tex]t=\frac{x}{2}[/tex].
Let's plug this into equation 2:
[tex]y=4(\frac{x}{2})^2+4(\frac{x}{2})-3[/tex]
[tex]y=4(\frac{x^2}{4})+2x-3[/tex]
[tex]y=x^2+2x-3[/tex]
This is the desired result.
Choice 3:
[tex]x=t-3[/tex]
[tex]y=t^2+2t[/tex]
Solve the first equation for [tex]t[/tex] by adding 3 on both sides:
[tex]x+3=t[/tex].
Plug into second equation:
[tex]y=(x+3)^2+2(x+3)[/tex]
Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:
[tex]y=(x^2+6x+9)+(2x+6)[/tex]
[tex]y=(x^2)+(6x+2x)+(9+6)[/tex]
[tex]y=x^2+8x+15[/tex]
Not the desired result.
Choice 4:
[tex]x=t^2[/tex]
[tex]y=2t-3[/tex]
I'm going to solve the bottom equation for [tex]t[/tex] since I don't want to deal with square roots.
Add 3 on both sides:
[tex]y+3=2t[/tex]
Divide both sides by 2:
[tex]\frac{y+3}{2}=t[/tex]
Plug into equation 1:
[tex]x=(\frac{y+3}{2})^2[/tex]
This is not the desired result because the [tex]y[/tex] variable will be squared now instead of the [tex]x[/tex] variable.
Choice 5:
[tex]x=t+1[/tex]
[tex]y=t^2+4t[/tex]
Solve the first equation for [tex]t[/tex] by subtracting 1 on both sides:
[tex]x-1=t[/tex].
Plug into equation 2:
[tex]y=(x-1)^2+4(x-1)[/tex]
Distribute and use the binomial square identity used earlier:
[tex]y=(x^2-2x+1)+(4x-4)[/tex]
[tex]y=(x^2)+(-2x+4x)+(1-4)[/tex]
[tex]y=x^2+2x+-3[/tex]
[tex]y=x^2+2x-3[/tex].
This is the desired result.
Answer:
Option 2; Option 5
Step-by-step explanation:
x² + 2x - 3 = 0
x = 2t
y = (2t)² +2(2t) - 3 = 4t² + 4t - 3
x = t+1
y = (t+1)² + 2(t+1) - 3
y = t² + 2t + 1 + 2t + 2 - 3 = t² + 4t
