Question 67286: 1.The sum of two positive real numbers is 7. The sum of their squares is 26.1/2. Find the numbers.
2.The perimeter of this right-angled triangle is 60 ( the two perpendicular sides are x and y and the other side is 26).
a) Write down two equations in x and y.
b) Find the values of x and y, given that x>y.
3.The diagram shows two squares; the sum of their areas is 58 meter square. The sum of the lengths of their sides is 10 meter. Find the values of x and y, as shown : (the small square is with the x side and the big square is with y side. the diagram shows that x+y = 10).
4. Solve:
5x-2y=3
xsquare+ysquare-2xy=0
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! These problems are similar. The Method
a) arrange the simplest equation to be used as substitution.
b) substitute in the 2nd equation, arrange as a quadratic equation
c) factor or solve the quadratic equation>
:
:
1.The sum of two positive real numbers is 7.
x + y = 7
y = 7 - x
:
The sum of their squares is 26.1/2. Find the numbers.
x^2 + y^2 = 26.5
;
Substitute (7-x) for y:
x^2 + (7-x)^2 = 26.5
x^2 + 49 - 14x + x^2 = 26.5
x^2 + x^2 - 14x + 49 - 26.5 = 0
2x^2 - 14x + 22.5 = 0
:
Simplify, divide by 2:
x^2 - 7x + 11.25 = 0
:
Solving this we get:
x = +2.5, x = +4.5 and these are to two numbers
:
:
2.The perimeter of this right-angled triangle is 60 ( the two perpendicular sides are x and y and the other side is 26).
a) Write down two equations in x and y.
:
x + y + 26 = 60
x + y = 60 - 26
x + y = 34
y = (34-x)
and
x^2 + y^2 = 26^2
x^2 + y^2 = 676
:
b) Find the values of x and y, given that x>y.
Substitute (34-x) for y in x^2 + y^ = 676:
x^2 + (34-x)^2 = 676
x^2 + 1156 - 68x + x^2 = 676
x^2 + x^2- 68x + 1156 - 676 = 0
2x^2 - 68x + 480 = 0
Simplify, divide by 2:
x^2 - 34x + 240 = 0
Factors to:
(x-10)(x-24) = 0
:
In the problem x = 24, y = 10
:
:
3.The diagram shows two squares; the sum of their areas is 58 meter square. The sum of the lengths of their sides is 10 meter. Find the values of x and y, as shown : (the small square is with the x side and the big square is with y side. the diagram shows that x+y = 10).
:
This is the same as the previous problems,
y = (10+x)
;
Substitute into x^2 + y^2 = 58 in the same way, solve the quadratic equation
:
:
4. Solve:
5x-2y=3
5x = 2y + 3
x = (2/5)y + (3/5)
x = .4y + .6 in decimals
:
x^2 + y^2 - 2xy = 0
:
Substitute (.4y + .6) for x:
(.4y+.6)^2 + y^2 - 2y(.4y+.6) = 0
.16y^2 + .48y + .36 + y^2 -.8y^2 - 1.2y = 0
.16y^2 + y^2 -.8y^2 + .48y - 1.2y + .36 = 0
.36y^2 - .72y + .36 = 0
:
Simplify, divide equation by .36, we have a simple equation:
y^2 - 2y + 1 = 0
Factors to:
(y - 1)(y - 1) = 0
y = +1
:
Find x:
5x - 2y = 3
5x - 2(1) = 3
5x = 3 + 2
x = 5/5
x = 1
:
Check: 1^2 + 1^2 - 2(1*1) = 0
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