Exercise 39 Page 590

We are given that the sum of three numbers is 180 and two of them are the same. Let's call each of them x. Additionally, we know that each of x is 13 of the greatest number, which we can call y. Using this information we can write a system of equations. 2 x+ y=180 & (I) x=13 y & (II) Let's solve this system using the Substitution Method. To do this, we will substitute 13y for x into the first equation.

2x+y=180 & (I) x= 13y & (II)

Substitute

(I):x= 13

2( 13y)+y=180 x= 13y

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(I):Solve for y

MoveLeftFacToNum

(I):a*b/c= a* b/c

23y+y=180 x= 13y

NumberToFrac

(I):a = 3* a/3

23y+ 33y=180 x= 13y

AddFrac

(I):Add fractions

53y=180 x= 13y

MultEqn

(I):LHS * 3/5=RHS* 3/5

35* 53y= 35*180 x= 13y

MultFracByInverse

(I):a/b* b/a=1

y= 35*180 x= 13y

MoveRightFacToNum

(I):a/c* b = a* b/c

y= 3*1805 x= 13y

ReduceFrac

(I):a/b=.a /5./.b /5.

y=3*36 x= 13y

Multiply

(I):Multiply

y=108 x= 13y

The greatest number is 108. To find the least number, we will substitute 108 for y in the second equation and evaluate.

y=108 x= 13y

Substitute

(II):y= 108

y=108 x= 13( 108)

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(II):Solve for x

MoveRightFacToNum

(II):a/c* b = a* b/c

y=108 x= 1*1083

ReduceFrac

(II):a/b=.a /3./.b /3.

y=108 x= 1*361

MultByOne

(II):a * 1=a

y=108 x= 361

DivByOne

(II):a/1=a

y=108 x=36

The least number is 36. This corresponds with answer C.