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2 years ago

Solve by using substitution x=3y-1 x+2y=9

Mathematics

1 answer:

kogti [31]2 years ago

3 0

Answer:

x=3y−1x=3y-1

x+2y=9x+2y=9

Replace all occurrences of xx with 3y−13y-1 in each equation.

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Replace all occurrences of xx in x+2y=9x+2y=9 with 3y−13y-1.

(3y−1)+2y=9(3y-1)+2y=9

x=3y−1x=3y-1

Add 3y3y and 2y2y.

5y−1=95y-1=9

x=3y−1x=3y-1

Solve for yy in the first equation.

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Move all terms not containing yy to the right side of the equation.

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Add 11 to both sides of the equation.

5y=9+15y=9+1

x=3y−1x=3y-1

Add 99 and 11.

5y=105y=10

x=3y−1x=3y-1

Divide each term by 55 and simplify.

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Divide each term in 5y=105y=10 by 55.

5y5=1055y5=105

x=3y−1x=3y-1

Cancel the common factor of 55.

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Cancel the common factor.

5y5=1055y5=105

x=3y−1x=3y-1

Divide yy by 11.

x=5,y=2

Step-by-step explanation:

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$\theta_{CAB}=128.316$

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$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

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the angle CAB is opposite to the BC.

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to find the CAB, we'll use:

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Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

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$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

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finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$

$25.842+128.316+25.842$

$180$

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