x=2+y
x^2=4+4y+y^2
so 4+4y+y^2+y^2=34
y^2+2y-30=0
y=(-2+-sqrt(4+120))/1=-1+- sqrt31
Put each of those y solutions into either equation, and solve for possible x, then test the solutions.
x² + y² = 34
x – y = 2
x^2=4+4y+y^2
so 4+4y+y^2+y^2=34
y^2+2y-30=0
y=(-2+-sqrt(4+120))/1=-1+- sqrt31
Put each of those y solutions into either equation, and solve for possible x, then test the solutions.
Hmmm. 3^2+5^2 = 34
Let's start by solving the second equation for x:
x - y = 2
If we add y to both sides of the equation, we get:
x = y + 2
Now, let's substitute this value of x into the first equation:
(y + 2)² + y² = 34
Expanding the squared term, we have:
(y² + 4y + 4) + y² = 34
Combining like terms, we get:
2y² + 4y + 4 = 34
Subtracting 34 from both sides, we have:
2y² + 4y - 30 = 0
Now, let's divide the entire equation by 2 to simplify it:
y² + 2y - 15 = 0
Factoring this equation, we have:
(y + 5)(y - 3) = 0
This gives us two possible values for y:
y + 5 = 0 or y - 3 = 0
Solving each equation separately, we find that:
y = -5 or y = 3
Now, let's substitute the value of y back into the equation x = y + 2:
If y = -5:
x = -5 + 2 = -3
If y = 3:
x = 3 + 2 = 5
Therefore, the solutions to the simultaneous equations are:
x = -3, y = -5
x = 5, y = 3
I hope my calculations didn't make your head spin!
Step 1: Rearrange the second equation to solve for x in terms of y.
The second equation is: x - y = 2.
Add y to both sides of the equation: x = y + 2.
Step 2: Substitute the expression for x in the first equation.
The first equation is: x² + y² = 34.
Substitute the value of x from the rearranged second equation into the first equation:
(y + 2)² + y² = 34.
Step 3: Expand and simplify the equation.
(y + 2)² = (y + 2)(y + 2) = y² + 4y + 4.
Substitute this expression back into the equation:
y² + 4y + 4 + y² = 34.
Step 4: Combine like terms and set the equation to zero.
2y² + 4y - 30 = 0.
Step 5: Solve the quadratic equation.
Use the quadratic formula to solve for y:
y = (-b ± √(b² - 4ac)) / (2a).
For the equation 2y² + 4y - 30 = 0, a = 2, b = 4, and c = -30.
Substitute these values into the quadratic formula:
y = (-4 ± √(4² - 4(2)(-30))) / (2(2)).
Simplify:
y = (-4 ± √(16 + 240)) / 4.
y = (-4 ± √256) / 4.
y = (-4 ± 16) / 4.
Step 6: Solve for y.
First, consider the positive solution:
y = (-4 + 16) / 4.
y = 12 / 4.
y = 3.
Next, consider the negative solution:
y = (-4 - 16) / 4.
y = -20 / 4.
y = -5.
Step 7: Find the corresponding values of x.
Using the expression we found in step 1:
For y = 3, x = 3 + 2 = 5.
For y = -5, x = -5 + 2 = -3.
Therefore, the solutions to the simultaneous equations are:
x = 5, y = 3
x = -3, y = -5