Question

Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).

Answer 1

The pairs (x,y) of real numbers such that x + y = 10 and x² + y² = 56 are (3.3, 6.7) and (6.7, 3.3)

How to determine the solution to the system of equations

From the question, we have the following parameters that can be used in our computation:

x + y = 10

y = 10 - x

Also, we have

x² + y² = 56

Substitute y = 10 - x into the equation x² + y² = 56

So, we have

x² + (10 - x)² = 56

Expand

x² + 100 + x² - 20x = 56

This gives

2x² - 20x + 100 - 56 = 0

2x² - 20x + 44 = 0

Divide through by 2

x² - 10x + 22 = 0

Solving graphically, we have

x = 3.3 and 6,7

Recall that

y = 10 - x

So, we have

y = 10 - 3.3 and y = 10 - 6.7

y = 6.7 and y = 3.3

So, the ordered pairs are (3.3, 6.7) and (6.7, 3.3)

Answer 2

(3.26795, 6.73205)

(6.73205, 3.26795)

Explanation:

Easiest and fastest way to get your solutions is to graph the systems of equations and analyze the graph for where they intersect.