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3 years ago

13

What is the solution to the system of equations below? HELP!!!! y = negative one-fourth x + 2 and 3 y = negative three-fourths x

minus 6 no solution infinitely many solutions (–16, 6) (–16, –2)

2 answers:

Oxana [17]3 years ago

5 0

Answer:

No solution

Step-by-step explanation:

Step 1: Write out equations

y = -1/4x + 2

3y = -3/4x - 6

Step 2: Substitution

3(-1/4x + 2) = -3/4x - 6

Step 3: Distribute

-3/4x + 6 = -3/4x - 6

From here, we can see that we have the same slope but different y-intercept. This means that the 2 lines are parallel and therefore never intersect.

Alternatively, you could graph the equations and see that the 2 lines are parallel and never intersect.

Goryan [66]3 years ago

4 0

Answer:

No solution

Step-by-step explanation:

y = -1/4x + 2

3y = -3/4x - 6

Plug y as -1/4x + 2 in the second equation.

3(-1/4x + 2) = -3/4x - 6

-3/4x + 6 = -3/4x - 6

-3/4x + 3/4x = -6 -6

0 = -12

No solution.

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Answer:

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Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

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Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

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Split

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$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

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$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

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$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

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When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

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$(R,W) \to (4000,0)$

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So, we have:

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