Solve each system of equations by substitution:
x^2 - y +2 = 0
4x = 14 -y
y + 2x = x^2 - 6
x + y - 3 = 2x^2
x^2 - y +2 = 0
4x = 14 -y
y=14-4x
x^2-(14-4x) +2=0
x^2+4x-12=0
(x+6)(x-2)=0
x=6; x=2, y=-10;6
To solve each system of equations by substitution, follow these steps:
1. Solve one equation for one variable.
2. Substitute the expression from step 1 into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value obtained in step 3 back into the equation from step 1 to solve for the first variable.
5. Check that the solutions obtained in steps 3 and 4 satisfy both original equations.
Let's solve the first system of equations:
1. From the first equation, solve for y:
x^2 - y + 2 = 0
Rearrange the equation:
y = x^2 + 2
2. Substitute y = x^2 + 2 into the second equation:
4x = 14 - y
Replace y with x^2 + 2:
4x = 14 - (x^2 + 2)
3. Solve the resulting equation for x:
4x = 14 - x^2 - 2
Rearrange the equation:
x^2 + 4x - 12 = 0
Factor the quadratic equation:
(x + 6)(x - 2) = 0
Solve for x:
x + 6 = 0 or x - 2 = 0
x = -6 or x = 2
4. Substitute the values of x back into y = x^2 + 2 to solve for y:
For x = -6:
y = (-6)^2 + 2
y = 36 + 2
y = 38
For x = 2:
y = (2)^2 + 2
y = 4 + 2
y = 6
5. Check the solutions by substituting them into the original equations:
For x = -6 and y = 38:
x^2 - y + 2 = 0
(-6)^2 - 38 + 2 = 0
36 - 38 + 2 = 0
0 = 0 (true)
4x = 14 - y
4(-6) = 14 - 38
-24 = -24 (true)
For x = 2 and y = 6:
y + 2x = x^2 - 6
6 + 2(2) = (2)^2 - 6
6 + 4 = 4 - 6
10 = -2 (false)
x + y - 3 = 2x^2
2 + 6 - 3 = 2(2)^2
5 = 8 (false)
Therefore, the solution to the first system of equations is x = -6 and y = 38.
Now, let's solve the second system of equations:
1. From the first equation, solve for y:
y = x^2 - 2x + 6
2. Substitute y = x^2 - 2x + 6 into the second equation:
x + y - 3 = 2x^2
Replace y with x^2 - 2x + 6:
x + (x^2 - 2x + 6) - 3 = 2x^2
3. Solve the resulting equation for x:
x + x^2 - 2x + 6 - 3 = 2x^2
Rearrange the equation:
x^2 - x - 3 = 0
Factor the quadratic equation:
(x - 3)(x + 1) = 0
Solve for x:
x - 3 = 0 or x + 1 = 0
x = 3 or x = -1
4. Substitute the values of x back into y = x^2 - 2x + 6 to solve for y:
For x = 3:
y = (3)^2 - 2(3) + 6
y = 9 - 6 + 6
y = 9
For x = -1:
y = (-1)^2 - 2(-1) + 6
y = 1 + 2 + 6
y = 9
5. Check the solutions by substituting them into the original equations:
For x = 3 and y = 9:
y + 2x = x^2 - 6
9 + 2(3) = (3)^2 - 6
9 + 6 = 9 - 6
15 = 3 (false)
x + y - 3 = 2x^2
3 + 9 - 3 = 2(3)^2
9 = 18 (false)
For x = -1 and y = 9:
y + 2x = x^2 - 6
9 + 2(-1) = (-1)^2 - 6
9 - 2 = 1 - 6
7 = -5 (false)
x + y - 3 = 2x^2
-1 + 9 - 3 = 2(-1)^2
5 = -2 (false)
Therefore, the second system of equations has no solution.