Question 67286
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>
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1.The sum of two positive real numbers is 7.
x + y = 7
y = 7 - x 
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The sum of their squares is 26.1/2. Find the numbers.
x^2 + y^2 = 26.5
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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
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Simplify, divide by 2:
x^2 - 7x + 11.25 = 0
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Solving this we get:
x = +2.5, x = +4.5 and these are to two numbers
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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.
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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
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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
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In the problem x = 24, y = 10
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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).
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This is the same as the previous problems,
y = (10+x)
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Substitute into x^2 + y^2 = 58 in the same way, solve the quadratic equation
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4. Solve:
5x-2y=3
5x = 2y + 3
x = (2/5)y + (3/5)
x = .4y + .6 in decimals
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x^2 + y^2 - 2xy = 0
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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
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Simplify, divide equation by .36, we have a simple equation:
y^2 - 2y + 1 = 0
Factors to:
(y - 1)(y - 1) = 0
y = +1
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Find x:
5x - 2y = 3
5x - 2(1) = 3
5x = 3 + 2
x = 5/5
x = 1
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Check: 1^2 + 1^2 - 2(1*1) = 0