Answer:
13.31 and - 3.31
Step-by-step explanation:
(x + y)^2 = x^2 + y^2 + 2xy ; x + y = 10; x^2 + y^2 = 56
This means ;
10^2 = 56 + xy;
100-56 = XY
44 = xy---------(3)
From x + y = 10
x = 10- y
Substitute that in eqn 3;
We have: 44 = (10-y) y
44 =10y - y^2
Y^2-10y -44 = 0;
From formula method of quadratic equation;
Y =[ - (-10) +_ √ (-10)^2 + 4 × 1 x (-44)] / 2 × 1
y = [ 10 +_√ (100 + 176) ]/ 2]
y = [ 10 +_ √ 276]/ 2
y = [ 10 + √ 276]/ 2 or [ 10 _ √ 276]/ 2
y = 13.31 or - 3.31
Similarly x = 10 - 13.31 or 10 - (-3.31)
x = - 3.31 or 13.31
The required ordered pairs are (5+√3, 5-√3) and (5-√3, 5+√3).
Given equations are:
[tex]x+y=10[/tex]......(1)
[tex]x^{2} +y^{2} =56[/tex].......(2)
An equation is a statement that equates to two statements.
On squaring equation (1) we get
[tex]x^{2} +y^{2} +2xy=100[/tex]
From equation(2) [tex]x^{2} +y^{2} =56[/tex]
So, [tex]2xy=44[/tex]
[tex]xy=22[/tex]
[tex]y=\frac{22}{x}[/tex]
Put [tex]y=\frac{22}{x}[/tex] in equation (1)
We get [tex]x+\frac{22}{x} =10[/tex]
[tex]x^{2} -10x+22=0[/tex]
[tex]x=5+\sqrt{3}[/tex]
[tex]x=5-\sqrt{3}[/tex]
If [tex]x=5+\sqrt{3}[/tex],
[tex]y=10-x[/tex]
[tex]y=5-\sqrt{3}[/tex]
So the ordered pair is [tex](5+\sqrt{3}, 5-\sqrt{3} )[/tex]
If [tex]x=5-\sqrt{3}[/tex],
[tex]y=10-x[/tex]
[tex]y=5+\sqrt{3}[/tex]
So the ordered pair is [tex](5-\sqrt{3}, 5+\sqrt{3} )[/tex]
Hence, the required ordered pairs are [tex](5+\sqrt{3}, 5-\sqrt{3} )[/tex] and [tex](5-\sqrt{3}, 5+\sqrt{3} )[/tex].
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