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# WHAT IS ?

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(1999 x 1/24 + ?) x 1 3/5 - 3 5/11 x 3 2/3 = 127

what is ?

Aug 17, 2018

#1
+2

Solve for M:
(8 (M + 1999/24))/5 - 38/3 = 127

Put each term in M + 1999/24 over the common denominator 24: M + 1999/24 = (24 M)/24 + 1999/24:
8/5 (24 M)/24 + 1999/24 - 38/3 = 127

(24 M)/24 + 1999/24 = (24 M + 1999)/24:
8/5 (24 M + 1999)/24 - 38/3 = 127

8/24 = 8/(8×3) = 1/3:
(24 M + 1999)/(3×5) - 38/3 = 127

3×5 = 15:
(24 M + 1999)/15 - 38/3 = 127

Put each term in (24 M + 1999)/15 - 38/3 over the common denominator 15: (24 M + 1999)/15 - 38/3 = (24 M + 1999)/15 - 190/15:
(24 M + 1999)/15 - 190/15 = 127

(24 M + 1999)/15 - 190/15 = ((24 M + 1999) - 190)/15:
(24 M - 190 + 1999)/15 = 127

Add like terms. 1999 - 190 = 1809:
(24 M + 1809)/15 = 127

Multiply both sides of (24 M + 1809)/15 = 127 by 15:
(15 (24 M + 1809))/15 = 15×127

(15 (24 M + 1809))/15 = 15/15×(24 M + 1809) = 24 M + 1809:
24 M + 1809 = 15×127

15×127 = 1905:
24 M + 1809 = 1905

Subtract 1809 from both sides:
24 M + (1809 - 1809) = 1905 - 1809

1809 - 1809 = 0:
24 M = 1905 - 1809

1905 - 1809 = 96:
24 M = 96

Divide both sides of 24 M = 96 by 24:
(24 M)/24 = 96/24

24/24 = 1:
M = 96/24

The gcd of 96 and 24 is 24, so 96/24 = (24×4)/(24×1) = 24/24×4 = 4:
M = 4

Aug 17, 2018
#2
+1

Hey guest,

Let's try and solve for the question mark. Note that I have replaced the question mark with a more standard symbol, x. Below is a representation of the equation in LaTeX form:

$$1\frac{3}{5}(1999*\frac{1}{24}+x)-3\frac{5}{11}*3\frac{2}{3}=127$$

 $$1\frac{3}{5}(1999*\frac{1}{24}+x)-3\frac{5}{11}*3\frac{2}{3}=127$$ Mixed numbers may be nice to look at, but they are fairly clumsy when algebraic computation is involved. I will convert all mixed numbers into improper fractions. $$\frac{8}{5}(\frac{1999}{24}+x)-\frac{38}{11}*\frac{11}{3}=127$$ Let's do some simplification and distribution. $$\frac{8}{5}*\frac{1999}{24}=\frac{1}{5}*\frac{1999}{3}=\textcolor{red}{\frac{1999}{15}}\\ \frac{38}{11}*\frac{11}{3}=38*\frac{1}{3}=\textcolor{blue}{\frac{38}{3}}$$ $$\textcolor{red}{\frac{1999}{15}}+\frac{8}{5}x-\textcolor{blue}{\frac{38}{3}}=127$$ In general, I recommend for situations like this to multiply both sides by the LCD. This will eliminate all instances of fractions. The LCD, in this case, is 15. $$1999+24x-190=1905$$ Let's do some more simplification on the left-hand side of the equation. $$24x+1809=1905$$ Subtract 1809 on both sides. $$24x=96$$ Divide by 24 on both sides to isolate x completely. $$x=4$$
Aug 17, 2018
#3
+1

Sorry X^2 !. I didn't see you working on it. Otherwise, I wouldn't have touched it, knowing how thorough you are.

Aug 17, 2018
#4
+1

Thanks for the compliment, Guest! I always strive to make my responses thorough while maintaining its comprehensibility.

I would not apologize. Your intent was clear: to serve the community well. You accomplished that goal, and you should receive an equal amount of commendment. It is also sometimes nice to view different approaches to the same problem.

TheXSquaredFactor  Aug 18, 2018
edited by TheXSquaredFactor  Aug 18, 2018
#5
+1

THANKS GUYS

p.s. I am grade 6... I cant understand that

Aug 19, 2018
#6
0

Thanks for your thanks and for your feedback.  This is always appreciated.

However,

Xsquaredfactor has explained this well.

This question is very difficult for someone in 6th class.

If you understand fractions you can understand his explanation.

Thanks Xsquared and answering guest.

Melody  Aug 19, 2018