how to solve

y= (x-2)^2+4
4x+2y=14
algebraically

By substitution:

y= (x-2)^2+4 ....(1)
4x+2y=14 ....(2)
Solve for y from (2) divided by 2:
2x+y=7
y=7-2x ...(2a)

Substitute (2a) in (1)
7-2x = (x-2)^2+4
Expand
7-2x = x^2-4x+4+4
x^2 -2x +1 = 0 ....(1a)

Solve 1(a) to get:
x=1 (duplicity 2)

Now substitute x=1 in (2a) to get
y=5

Substitute x=1,y=5 in (1) to check:
(1-2)^2+4 = 5 OK.

To solve the system of equations algebraically, we can use the method of substitution. Here are the step-by-step instructions:

Step 1: Solve one equation for one variable in terms of the other variable.
Let's solve the first equation, y = (x-2)^2 + 4, for y:
y = x^2 - 4x + 4 + 4
Simplifying, we have:
y = x^2 - 4x + 8

Step 2: Substitute the expression for the solved variable into the other equation.
Using the expression for y in the second equation, 4x + 2y = 14, we can substitute y:
4x + 2(x^2 - 4x + 8) = 14

Step 3: Simplify and solve for x.
Expanding and simplifying the equation, we get:
4x + 2x^2 - 8x + 16 = 14
2x^2 - 12x + 16 = 14
2x^2 - 12x + 2 = 0

Step 4: Factor or use the quadratic formula to solve for x.
To solve the quadratic equation, let's first divide the entire equation by 2 to simplify it:
x^2 - 6x + 1 = 0

Unfortunately, the equation cannot be easily factored, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = 1, b = -6, and c = 1. Substituting these values into the formula, we have:
x = (-(-6) ± √((-6)^2 - 4 * 1 * 1))/(2 * 1)
x = (6 ± √(36 - 4))/(2)
x = (6 ± √(32))/2
x = (6 ± 4√2)/2

Breaking down further, we have:
x = (6/2) ± (4√2)/2
x = 3 ± 2√2

So the solutions for x are x = 3 + 2√2 and x = 3 - 2√2.

Step 5: Substitute the values of x back into one of the original equations to solve for y.
Let's use the equation y = x^2 - 4x + 8 to find the corresponding y-values for each x-value:
For x = 3 + 2√2:
y = (3 + 2√2)^2 - 4(3 + 2√2) + 8
y = 9 + 12√2 + 8 - 12 - 8√2 + 8
y = 17 + 12√2 - 8√2

For x = 3 - 2√2:
y = (3 - 2√2)^2 - 4(3 - 2√2) + 8
y = 9 - 12√2 + 8 - 12 + 8√2 + 8
y = 17 - 12√2 + 8√2

So the solutions for y are:
y = 17 + 12√2 - 8√2
y = 17 - 12√2 + 8√2

Therefore, the solution to the system of equations is:
(x, y) = (3 + 2√2, 17 + 12√2 - 8√2) and (x, y) = (3 - 2√2, 17 - 12√2 + 8√2).

To solve the system of equations algebraically, we can use the method of substitution or elimination. Let's use the substitution method.

Step 1: Solve one equation for one variable in terms of the other variable.
We have the equation of the parabola:

y = (x-2)^2 + 4

Step 2: Substitute the expression for the variable found in step 1 into the other equation.
Substitute y = (x-2)^2 + 4 into the second equation:

4x + 2((x-2)^2 + 4) = 14

Step 3: Simplify and solve for x.
Start by expanding the squared term:

4x + 2(x^2 - 4x + 4) + 8 = 14
4x + 2x^2 - 8x + 8 + 8 = 14
2x^2 - 4x + 16 = 14

Rearrange to set the equation equal to zero:

2x^2 - 4x + 2 = 0

Step 4: Solve the quadratic equation.
We can solve it by factoring, completing the square, or using the quadratic formula.

Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values:

x = (-(-4) ± √((-4)^2 - 4(2)(2))) / (2(2))
x = (4 ± √(16 - 16)) / 4
x = (4 ± √0) / 4
x = (4 ± 0) / 4

There is a double root, x = 1.

Step 5: Substitute the value of x into either of the original equations to solve for y.
Substitute x = 1 into the first equation:

y = (1-2)^2 + 4
y = (-1)^2 + 4
y = 1 + 4
y = 5

Therefore, the solution to the system of equations is x = 1 and y = 5.