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pshichka [43]
3 years ago
13

Problem 1 3x + y = –14 -2x-y=9

Mathematics
1 answer:
Radda [10]3 years ago
3 0

Answer:

x = -5 , y = 1

Step-by-step explanation:

Solve the following system:

{3 x + y = -14 | (equation 1)

-2 x - y = 9 | (equation 2)

Add 2/3 × (equation 1) to equation 2:

{3 x + y = -14 | (equation 1)

0 x - y/3 = (-1)/3 | (equation 2)

Multiply equation 2 by -3:

{3 x + y = -14 | (equation 1)

0 x+y = 1 | (equation 2)

Subtract equation 2 from equation 1:

{3 x+0 y = -15 | (equation 1)

0 x+y = 1 | (equation 2)

Divide equation 1 by 3:

{x+0 y = -5 | (equation 1)

0 x+y = 1 | (equation 2)

Collect results:

Answer: {x = -5 , y = 1

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Step-by-step explanation:

Original quadratic equation is 9x^{2}-3x-2=0

Divide both sides by 9:

x^{2} - \frac{x}{3} - \frac{2}{9}=0

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x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

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x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

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\left(x - \frac{1}{6}\right)^2=\frac{1}{4}

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\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

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x - \frac{1}{6}=-\sqrt{\frac{1}{4}}  and x - \frac{1}{6}=\sqrt{\frac{1}{4}}

\sqrt{\frac{1}{4}} = \frac{1}{2}

So the two roots are

x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}

and

x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

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