Useing substitution/elimination method what is the solution set to 2x+y=1
x^2=y+2
a)(-3,7) and (1,-1)
b)(3,7) and (-1,-1)
c)(-3,-7) and (1,1)
d)(3,-7) and (-1,1)
I've really no educated guess so PLEASE HELP!
You can make an educated guess by substituting the various (x,y) combinations and eliminating the ones that don't work.
For example, if x = 1 and y = 1, obviously the requirement 2x + y = 1 is not satisfied. Therefore you can eliminate (c)
That is probably what they mean by the "substitution/elimination" method. It's only useful when you are presented with multiple choices.
To solve this system of equations using the substitution method, you'll start by solving one equation for one variable and then substituting it into the other equation.
1. Let's solve the first equation for y:
2x + y = 1
Subtract 2x from both sides:
y = 1 - 2x
2. Now substitute this expression for y into the second equation:
x^2 = y + 2
Replace y with 1 - 2x:
x^2 = (1 - 2x) + 2
3. Simplify the equation:
x^2 = 1 - 2x + 2
x^2 = 3 - 2x
4. Move all the terms to one side to get a quadratic equation:
x^2 + 2x - 3 = 0
Now, to find the values of x, you can either factor the quadratic equation or use the quadratic formula.
5. Factoring the quadratic equation:
(x + 3)(x - 1) = 0
Setting each factor equal to zero gives you two possible values for x:
x + 3 = 0 or x - 1 = 0
x = -3 or x = 1
6. Now substitute these values for x back into the first equation to find the values of y:
For x = -3:
2(-3) + y = 1
-6 + y = 1
y = 7
So one solution is (-3, 7).
For x = 1:
2(1) + y = 1
2 + y = 1
y = -1
So the other solution is (1, -1).
In conclusion, the solution set to the given system of equations is (a) (-3, 7) and (1, -1).