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kap26 [50]
3 years ago
5

All questions to be solved using linear combination.

Mathematics
1 answer:
monitta3 years ago
8 0
1)
I:x-y=-7
II:x+y=7

add both equations together to eliminate y:
x-y+(x+y)=-7+7
2x=0
x=0

insert x=0 into II:
0+y=7
y=7

the solution is (0,7)

2)
I: 3x+y=4
II: 2x+y=5

add I+(-1*II) together to eliminate y:
3x+y+(-2x-y)=4+(-5)
x=-1

insert x=-1 into I:
3*-1+y=4
y=7

the solution is (-1,7)

3)

I: 2e-3f=-9
II: e+3f=18

add both equations together to eliminate f:
2e-3f+(e+3f)=-9+18
3e=9
e=3

insert e=3 into I:
2*3-3f=-9
-3f=-9-6
-3f=-15
3f=15
f=5

the solution is (3,5)

4)
I: 3d-e=7
II: d+e=5

add both equations together to eliminate e:
3d-e+(d+e)=7+5
4d=12
d=3

insert d=3 into II:
3+e=5
e=2

the solution is (3,2)

5)
I: 8x+y=14
II: 3x+y=4

add I+(-1*II) together to eliminate y
8x+y+(-3x-y)=14-4
5x=10
x=2

insert x=2 into II:
3*2+y=4
y=4-6
y=-2

the solution is (2,-2)
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Andrew [12]

Answer:

10 jelly beans.

Step-by-step explanation:

Let x be ounces of jelly beans.

We have been given that a jar contains 10 ounces of gumdrops, 3 ounces of licorice bits, and 3 ounces of jelly beans.

To make weight of jelly beans be 50% of total assortment, we will add x ounces of jelly beans to 3 so that it will be 50% of total assortment plus x. We can represent this information in an equation as:

3+x=\frac{50}{100}(10+3+3+x)

Let us solve for x.

3+x=0.50(16+x)

3+x=8+0.5x

3-3+x=8-3+0.5x

x=5+0.5x

x-0.5x=5+0.5x-0.5x

0.5x=5

\frac{0.5x}{0.5}=\frac{5}{0.5}

x=10

Therefore, 10 jelly beans must be added to make the assortment 50 percent jelly beans in weight.

Let us cross check our answer.

\frac{3+10}{3+3+10+10}=\frac{13}{26}=0.5=50\%

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