**L**= larger number

**S**= smaller number

**L**+

**S**= 35 from which

**L**= 35 -

**S**or

**S**= 35 -

**L**

**L**= 4*

**S**

**S**) = 4*

**S**

**S**= 4*

**S**

**S**to each side

**S***(4 + 3) = 7

**S**

**S**= 15

**L**= 35 - S = 35 - 15 =

**20**

ok so i have to figure out how to write this down so i can solve it but i have no idea how to write that down in allgebraic form

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3***L** = 4***S**

Therefore 3*(35 - **S**) = 4***S**

105 - 3***S** = 4***S**

Add 3***S** to each side

105 = **S***(4 + 3) = 7**S**

Therefore **S** = 15

L+S=35

3L=4S

L+S=35

3L-4S=0

multiply 1sr equation by 3

3L+3S=105

3L-4S=0

subtract

7S=105

S=15

L=35-15=20

Both approaches are correct but a different one variable approach seem s simpler

larger number = L the other number = 35-L

3L=4(35-L)

3L=140-4L

7L=140

L=20

Hi Liz;

I like William's answer. However, I prefer to only work with one variable.

x=larger number

(3/4)x=smaller number

(x)+[(3/4)(x)]=35

[(4/4)(x)]+[(3/4)(x)]=35

(7/4)x=35

Let's multiply both sides by 4/7

(4/7)(7/4)x=35(4/7)

x=20

If you need the smaller number...

35-20=15

To verify...

Does

3(larger)=4(smaller)

3(20)=4(15)

60=60

YES!

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