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3 years ago

Solve the system by elimination. 3x-10y=-25 4x+40y=20

2 answers:

7 0

3x-10y=-25Add 10y10y to both sides
3x=−25+10y

Regroup terms
3x=10y-253x=10y−25
Divide both sides by 3 3
x=10y-25/3

Factor out the common term 5 5
your answer: x=5(2y-5)/3

4x+40y=20
Factor out the common term 4 4
4(x+10y)=20
Divide both sides by 44
x+10y=20/4

Simplify 20/4 to 5
x+10y=5

Subtract 10y10y from both sides
your answer: x=5-10y

lana66690 [7]3 years ago

3 0

3x-10y=-25
4x+40y=20

12x-40y=-100
12x+120y=60

-160y=40

y=1/4
x=5/2

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3 years ago

The pinata weighs 11/4 pounds without candy.He adds 8bags that weighs 3/8,3/8,1/2,1/2,1/2,3/4,3,11/4 how much will the pinata we

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Answer:

$Weight = \frac{23}{2}$

Step-by-step explanation:

Given

Weight of Pinata = 11/4 lb

Weight of candies = 3/8, 3/8, 1/2 , 1/2, 1/2, 3/4, 3, 11/4

Required

Weight of Pinata when filled with candies

To solve this question, we simply need to add all weights together

First, we need to add up the weight of the candies

$Candies = \frac{3}{8} + \frac{3}{8} + \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \frac{3}{4} + 3 + \frac{11}{4}$

Take LCM

$Candies = \frac{3 + 3 + 4 +4 + 4 + 6 + 24 + 22}{8}$

$Candies = \frac{70}{8}$

Simplify:

$Candies = \frac{35}{4}$

Next; Add the weight of the candies to the weight of pinata

$Weight = Candies + Pinata$

$Weight = \frac{35}{4} + \frac{11}{4}$

Take LCM

$Weight = \frac{35 + 11}{4}$

$Weight = \frac{46}{4}$

Simplify

$Weight = \frac{23}{2}$

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3 years ago

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Snezhnost [94]

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Step-by-step explanation:

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Answer:

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