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stira [4]
3 years ago
7

Solve the system of linear equations using elimination. −7x + 3y = −6 −3x + 3y = 6

Mathematics
2 answers:
Marysya12 [62]3 years ago
8 0

Answer:

x = 3 , y = 5

Step-by-step explanation:

Solve the following system:

{3 y - 7 x = -6 | (equation 1)

3 y - 3 x = 6 | (equation 2)

Subtract 3/7 × (equation 1) from equation 2:

{-(7 x) + 3 y = -6 | (equation 1)

0 x+(12 y)/7 = 60/7 | (equation 2)

Multiply equation 2 by 7/12:

{-(7 x) + 3 y = -6 | (equation 1)

0 x+y = 5 | (equation 2)

Subtract 3 × (equation 2) from equation 1:

{-(7 x)+0 y = -21 | (equation 1)

0 x+y = 5 | (equation 2)

Divide equation 1 by -7:

{x+0 y = 3 | (equation 1)

0 x+y = 5 | (equation 2)

Collect results:

Answer:  {x = 3 , y = 5

jolli1 [7]3 years ago
3 0

Answer:

(3, 5)

Step-by-step explanation:

−7x + 3y = −6

−3x + 3y = 6

Subtract the second equation from the first.

−4x = −12

x = 3

Then substitute. → −7x + 3y = −6 → −7(3) + 3y = −6

y = 5

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George and Riley are selling fruit for a school fundraiser. Customers can buy small boxes of oranges and large boxes of oranges.
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The cost of small box of oranges is $7.

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<u>Step-by-step explanation:</u>

It is given that,

3 small boxes of oranges and 14 large boxes of oranges for a total of $203.

11 small boxes of oranges and 11 large boxes of oranges for a total of $220.

Let us take,

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  • The cost of large box of oranges = y

<u>The system of equations are framed as :</u>

3x + 14y = 203  ---------(1)

11x + 11y = 220  ---------(2)

<u>To solve these equations for x and y values :</u>

Multiply equation (1) by 11 and equation (2) by 3

Subtract eq(2) from eq(1),

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⇒ y = 1573 / 121

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∴ The cost of large box of oranges is $13.

Substitute y=13 in the eq(1),

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⇒ 3x = 203 - 182

⇒ 3x = 21

⇒ x = 21 / 3

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∴ The cost of small box of oranges is $7.

<u />

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