Question 1134171
let x = ?


you can solve these as ratio problems by cross multiplying and then solving for x.


first problem is 7/12 = x/48
cross multiply to get 7 * 48 = 12 * x
divide both sides by 12 to get 7 * 48 / 12 = x
solve for x to get x = 28
first solution is 7/12 = 28/48


second problem is -6/b = x/4b
cross multiply to get -6 * 4b = b * x
simplify to get -24 * b = b * x
divide both sides by b to get -24 = x
solve for x to get x = -24
second solution is -6/b = -24/4b


third problem is 6/(5t^5*u^2) = x/(10t^7*u^5)
cross multiply to get 6 * 10t^7 * u^5 = 5t^5 * u^2 * x
divide both sides by (5t^5 * u^2) to get 6 * 10t^7 *u^5 / (5t^5 * u^2) = x
solve for x to get x = 6 * 2t^2 * u^3
third solution is 6/(5t^5*u^2) = 6*2t^2*u^3/(10t^7*u^5)
this can be simplified to 6/(5t^5*u^2) = 12*t^2*u^3 / (10t^7*u^5)
third solution is therefore 6/(5t^5*u^2) = 12*t^2*u^3 / (10t^7*u^5)


fourth problem is 8/(m-8) = x/(m*(m-8))
cross multiply to get 8 * m * (m-8) = (m - 8) * x
divide both sides by (m-8) to get 8 * m * (m-8) / (m-8) = x
solve for x to get x = 8 * m
fourth solution is 8/(m-8) = 8m/(m*(m-8))


how do you know the solutions are correct?
one way is to cross multiply again.
if the cross multiplications are equal to each other, then the ratios are equivalent.


for problem 1, 7/12 = 28/48 becomes 7*48 = 12*28 which becomes 336 = 336.


for problem 2, -6/b = -24/4b becomes -6*4b = -24*b which becomes -24*b = -24*b.


i'll save problem 3 for last since that's the hardest.


for problem 4, 8/(m-8) = 8m/(m*(m-8) becomes 8*m*(m-8) = (m-8) * 8m which becomes 8 * m * (m-8) = 8 * m * (m-8).


for problem 3, 6/(5t^5*u^2) = 12*t^2*u^3/(10t^7*u^5) becomes 6 * 10 * t^7 * u^5 = 5 * t^5 * u^2 * 12 * t^2 * u^3


6 * 10 * t^7 * u^5 becomes 60 * t^7 * u^5.


5 * t^5 * u^2 * 12 * t^2 * u^3 becomes:
5 * 12 * t^5 * t^2 * u^2 * u^3 which becomes:
60 * t^7 * u^5


final answer for problem 3 is 60 * t^7 * u^5 = 60 * t^7 * u^5


all the cross multiplications are equal therefore all the ratios are the same therefore the solutions for x are all good.


since x represents ?, then all the solutions for ? are good.


for problem 3, a couple of concepts dealing with exponent arithmetic helped.


those are:


x^a * x^b = x^(a+b)
x^a / x^b = x^(a-b)


another way is to multiply the first ratio by the same factor in the numerator and denominator to get the second ratio.


for the first problem, 7/12 * 4/4 = 28/48.


for the second problem, -6/b * 4/4 = -24/4b.


for the fourth problem (again saving problem 3 for last), 8/(m-8) * m/m = 8m/(m(m-8).


for the third problem, 6/(5*t^5*u^2) * 2*t^2*u^3/(2*t^2*u^3) equals:
6*2*t^2*u^3 / (5*t^5*u^2 * 2*t^2*u^3) which equals:
12*t^2*u^3 / (10*t^7*u^5)


the laws of exponent arithmetic apply.


t^5 * t^2 = t^(5+2) = t^7
u^2 * u^3 = u^(2+3) = u^5


another way to prove the solution to problem 3 is correct is to assign random values to u and t and then evaluate.


for example:


when u = 3 and t = 4, 6/(5t^5*u^2) = 12*t^2*u^3 / (10t^7*u^5) becomes:
6/(5*4^5*3^2) = 12*4^2*3^3/(10*4^7*3^5) which becomes:
6/46080 = 5184/39813120.


5184/6 = 864
therefore, if the ratio is good, you should get:
6/46080 * 864/864 = 5184/39813120.
6*864 = 5184
46080 * 864 = 39813120.


this proves that 6/46080 is the same ratio as 5184/39813120.
this also confirms that the solution to problem 3 is correct.