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3 years ago

How do i solve the linear system -14x+15y=15 21x-20y=-10 using elimination

Mathematics

1 answer:

valina [46]3 years ago

4 0

Solve the following system using elimination:
{15 y - 14 x = 15 | (equation 1)
{21 x - 20 y = -10 | (equation 2)

Swap equation 1 with equation 2:
{21 x - 20 y = -10 | (equation 1)
{-(14 x) + 15 y = 15 | (equation 2)

Add 2/3 × (equation 1) to equation 2:
{21 x - 20 y = -10 | (equation 1)
{0 x+(5 y)/3 = 25/3 | (equation 2)

Multiply equation 2 by 3/5:
{21 x - 20 y = -10 | (equation 1)
{0 x+y = 5 | (equation 2)

Add 20 × (equation 2) to equation 1:
{21 x+0 y = 90 | (equation 1)
{0 x+y = 5 | (equation 2)

Divide equation 1 by 21:
{x+0 y = 30/7 | (equation 1)
{0 x+y = 5 | (equation 2)

Collect results:

Answer: {x = 30/7, y = 5

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tekilochka [14]

Answer:

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3 years ago

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3 years ago

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Answer:

u <= 80

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3 years ago

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malfutka [58]

Answer:

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Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

What is the midpoint of EC ?

sladkih [1.3K]

<u>Given</u>:

Given that the graph OACE.

The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)

We need to determine the midpoint of EC.

<u>Midpoint of EC:</u>

The midpoint of EC can be determined using the formula,

$Midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substituting the coordinates E(2t,0) and C(2p, 2r), we get;

$Midpoint=(\frac{2t+2p}{2},\frac{0+2r}{2})$

Simplifying, we get;

$Midpoint=(\frac{2(t+p)}{2},\frac{2r}{2})$

Dividing, we get;

$Midpoint=(t+p,r)$

Thus, the midpoint of EC is (t + p, r)

Hence, Option A is the correct answer.

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3 years ago