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Oksi-84 [34.3K]
4 years ago
5

In the solution to this system, what is the value of y? x+y+z=1x+y+z=1 2x+y−z=82x+y-z=8 x−y+z=−5

Mathematics
1 answer:
kirza4 [7]4 years ago
3 0
If you have multiple equations with multiple variables, you can either do clever substitutions, or turn it into a matrix on which you can perform linear combinations or multiplications (Gauss elimination)

1  1   1  1
2  1  -1  8
1 -1   1  -5

(note how the above 3 rows represent the 3 equations, just got rid of the variables, plus sign and equals sign)

subtract row1 from row3, that eliminates x and z from row 3.

1  1   1  1
2  1  -1  8
0 -2   0  -6

divide row3 by -2, that will give y a factor of 1

1  1   1  1
2  1  -1  8
0 1   0  3

The last row now says y=3



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3 years ago
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A school fair ticket costs $8 per adult and $1 per child. On a certain day, the total number of adults (a) and children (c) who
gtnhenbr [62]

Let a be the number of adults which bought a ticket and c the number of children who bought a ticket. We can use the information given to assemble a system of equations.

Tickets cost 8 dollars for a, 1 dollar for c, and the total amount made is 100

8a + c = 100

The total of a and c is 30

a + c = 30

We can now subtract one equation from the other to use the elimination method to solve the system.

8a + c = 100

-(a + c = 30)

7a + 0 = 70

We can now solve for a.

7a = 70

a = 10

There were 10 adults who bought tickets. We can use this value as a known, plugging it into an equation to solve for the number of children.

a + c = 30

10 + c = 30

c = 20

So, the final answer is "20 children and 10 adults Equation 1: a + c = 30 Equation 2: 8a + c = 100".

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