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____ [38]
3 years ago
13

Solve the system of equations using subtraction. Equation 1: 3x-2y= -2 Equation 2: 3x + y = 10 The solution is (? ,?) HURRY PLEA

SE!!!!
Mathematics
1 answer:
Art [367]3 years ago
5 0

Answer:

the solution is x = 2 and y = 4 OR (2,4)

Step-by-step explanation:

From the question

Equation 1: 3x-2y= -2

Equation 2: 3x + y = 10

Subtract equation 1 from equation 2, that is

3x + y = 10 - (3x -2y = -2)

You get

3x - 3x + y - (-2y) = 10 - (-2)

0 + y + 2y = 10 + 2

3y = 12

Divide both sides by 3

∴ y = 12/3

y = 4

Substitute the value of y into equation 2 to get x

3x + y = 10

3x + 4 = 10

Then,

3x = 10 - 4

3x = 6

Divide both sides by 3

∴ x = 6/3

x = 2

Hence, the solution is x = 2 and y = 4

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Step-by-step explanation:

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C = sqrt(205) = 14.32

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2 years ago
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The answer would be 1/45
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Find the value of x. Write your answer in simplest form.
slega [8]

Answer:

x = (9/2)√2

Step-by-step explanation:

The ratio of leg x to the hypotenuse 9 is equal to the cosine of 45 degrees:

cos 45 degrees = x/9

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2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One
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y'=(t+y)^2-1

Substitute u=t+y, so that u'=y', and

u'=u^2-1

which is separable as

\dfrac{u'}{u^2-1}=1

Integrate both sides with respect to t. For the integral on the left, first split into partial fractions:

\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1

\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt

\dfrac12(\ln|u-1|-\ln|u+1|)=t+C

Solve for u:

\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C

\ln\left|1-\dfrac2{u+1}\right|=2t+C

1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}

\dfrac2{u+1}=1-Ce^{2t}

\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}

u=\dfrac2{1-Ce^{2t}}-1

Replace u and solve for y:

t+y=\dfrac2{1-Ce^{2t}}-1

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Now use the given initial condition to solve for C:

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y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}

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