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# pls help

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Find all ordered pairs x, y of real numbers such that x+y=10 and x^2+y^2=64.

For example, to enter the solutions (2, 4) and (-3, 9), you would enter "(2,4),(-3,9)" (without the quotation marks).

Apr 25, 2024

#1
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We can solve for x and y in this system of equations by using substitution or the method of completing the square. Here, we'll use substitution:

Solve one equation for one variable:

From the first equation, we can express one variable in terms of the other:

x = 10 - y (Equation 1)

Substitute into the second equation:

Substitute this expression for x in the second equation:

(10 - y)^2 + y^2 = 64

Expand and solve for y:

Expand the squared term:

100 - 20y + y^2 + y^2 = 64 Combine like terms: 2y^2 - 20y + 36 = 0 Factor the equation: 2(y^2 - 10y + 18) = 0

This factors further as: 2(y - 6)(y - 3) = 0

Therefore, the possible values for y are:

y = 6 or y = 3

Substitute y back to find x:

Substitute each value of y back into Equation (1) to find the corresponding value of x:

If y = 6:

x = 10 - 6 = 4

If y = 3:

x = 10 - 3 = 7

Solutions:

Therefore, the ordered pairs (x, y) that satisfy the system of equations are:

(x, y) = (4, 6)

(x, y) = (7, 3)

May 9, 2024