Question 72980
1. Two kilograms of iron was melted with 7 kilograms of other metals to make the alloy. If 1440 kilograms of the alloy was required, how many kilograms of iron should be used?
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the total is 2 + 7 = 9: Relationship of iron to the total: 2:9
Let I = kilos of iron
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{{{2/9}}} ={{{I/1440}}}
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Cross multiply:
9I = 2(1440)
9I = 2880
I = 2880/9
I = 320 kilos of iron in 1440 kilos of alloy
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2. Charles and Matthew knew that the formula for sulfuric acid was H2SO. If they had 196 grams of sulfuric acid, what was the weight of the sulfur(H,1; S,32; O,16)?
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The relationship of a sulfur weight to the total: 32:(2+32+16) = 32:50
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{{{32/50}}} = {{{S/196}}}
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Cross multiply:
50S = 6272
s = 6272/50
s = 125.44 grams of sulfur in 196 grams of H2SO
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3. The ratio of the two numbers was 7 to 2. When Sir Richard and MArion multiplied the denominator by 10, they found that the result was 84 greater than twice the numerator. What were the numbers?
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"The ratio of the two numbers was 7 to 2"
Let x = the multiplier
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{{{(7x)/(2x)}}}
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"Multiply the denominator (2x) by 10. 
Result is 84 greater than twice the numerator (7x)"
10(2x) = 2(7x) + 84
20x = 14x + 84
20x - 14x = 84
6x = 84
x = 14 is the multiplier
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Therefore the numbers are: (14*7) and 14(2) or 98 and 28  
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4. Solve: 5x-2 over(/) 3 -x over(/) 4 = 7
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{{{((5x-2))/3}}} - {{{x/4}}} = 7
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Multiplying equation by 12 gets rid of the denominators:
4(5x-2) - 3(x) = 12(7)
20x - 8 - 3x = 84
20x - 3x = 84 + 8
17x = 92
x = 92/17
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5. 1 over (/) x+3 + 3x over(/) x+2 + 2x+1 over(/) x squared + 5x + 6
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The way I understand it:
{{{1/(x+3) + 3x/(x+2) + (2x+1)/(x^2 + 5x + 6)}}}

The last denominator can be factored to the same values as the other 2 denominators:
{{{1/(x+3) + 3x/(x+2) + (2x+1)/((x+2)(x+3))}}}
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(x+3)(x+2) is the common denominator
{{{(1(x+2) + 3x(x+3) + (2x+1))/((x+2)(x+3))}}} = {{{(x + 2 + 3x^2 + 9x + 2x + 1 )/((x+3)(x+2))}}} = {{{( 3x^2 + 12x + 3 )/((x+3)(x+2))}}} = {{{( 3(x^2 + 4x + 1 ))/((x+3)(x+2))}}}
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That's about as far as you can go with it