Solving system of equations

Question

Solve for all real numbers "x" and "y" that satisfy the following system of equations:

x² + y² = 25

x + y = 7

Answer 1

graph the circle and the line'

find where they intersect

at (4,3) and (3,4)

Answer 2

To solve the system of equations x² + y² = 25 and x + y = 7, we start by using the second equation to rewrite one variable in terms of the other. Since x + y = 7, we can say y = 7 - x. Now we take that expression and plug it into the first equation, replacing y with 7 - x. This gives us x² + (7 - x)² = 25. Next, we expand (7 - x)², which becomes 49 - 14x + x². Now the equation is x² + 49 - 14x + x² = 25. Combine the like terms to get 2x² - 14x + 49 = 25. Then subtract 25 from both sides: 2x^2 - 14x + 24 = 0. Divide everything by 2 to make it simpler: x² - 7x + 12 = 0. This is a quadratic equation, and it factors into (x - 3)(x - 4) = 0, so x = 3 or x = 4. Now go back to the second equation, x + y = 7, to find y. If x = 3, then y = 4, and if x = 4, then y = 3. So the two answers are (3, 4) and (4, 3).

Answer 3

We have a system of equations here

x² + y² = 25

x + y = 7

I want to denote y in terms of x, so that we can reduce the variable to 1.

Then, we get y = 7 - x

We substitute this to the right side and get

x² + (7-x)² = 25 Expanding this, we get

2x² -14x+49=25

2x² -14x+24=0

Dividing both sides by 2, we get

x² -7x+12=0

(x-3)(x-4)=0

We get x = 3 and x = 4, and consequently, y = 4 and y = 3.

(x,y) = (3, 4) or (4,3)

Answer 4

Another approach.

x²+y²=25 Eqn 1

x+y=7 Eqn 2

Square Eqn2

x²+2xy+y²=49

Subtract Eqn1 from this

2xy=24

xy=12 Eqn 3

From Eqn2, y=7-x and substituting into Eqn 3

x(7-x)=12

7x-x²=-12

x²-7x+12 =0

(x-3)(x-4)=0

x =3,4

From Eqn 2,

When x=3, y=4

When x=4, y=3

Answer 5

We have x² + y² = 25, let's call that equation (1)

and we can call x + y = 7 equation (2).

From equation (2) we can subtract y from both sides to get x = 7 - y, and substitute our new "x" into equation (1)!

We will get: (7 - y)² + y² = 25, we can expand (7 - y)² by the property (a - b)² = a² - 2ab + b²,

so (7 - y)² = 49 - 14y + y²

So now we have

49 - 14y + y² + y² = 25

Combining our y²'s we get 49 - 14y + 2y² = 25, now we can subtract 25 from both sides and get

24 - 14y + y² = 0

We rearrange by decreasing degrees, and our equation will now look like this:

2y² - 14y + 24 = 0.

It is important here to notice that all our coefficients are even, so we can factor out a 2! This is important because, even though it is not necessary, quadratic equations are easier to solve when the leading coefficient is 1.

So now we have 2(y² - 7y + 12) = 0, and since 0 is divisible by 2, we divide both sides by 2 and get the following:

y² - 7y + 12 = 0.

Now all we need to find is 2 numbers whose product is 12 and sum is equal to -7, so -3 and -4 work, making our quadratic factored into: (y - 3)(y - 4) = 0 giving us y = 3 or y = 4.

Alternatively, or if no 2 numbers are easily found, we can use the quadratic formula which states if we have a quadratic formula in the form of ax2 + bx + c = 0, where a, b and c are real numbers and x is our variable, our solutions for x are in the form of: x = (-b +/- sqrt(b² - 4ac))/2a, so in our case we get y = (7+/- sqrt(49 - 48))/2 so y = (7 + 1)/2 or (7 - 1)/2, giving us y = 3 or y = 4, same answer as before!

Using our new y-values and plugging them back into equation (2) we have x = 7 - y.

For y = 3: x = 7 - 3, so x = 4.

For y = 4: x = 7 - 4, so x = 3.

To make sure of our work we can plug our values back into equations (1) and (2) to see if they work!

4² + 3² = 16 + 9 = 25, this works.

and 4 + 3 = 7, this works as well!

Similarly we get 3² + 4² = 9 + 16 = 25, and 3 + 4 =7, so our solutions work.