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2 years ago

What is the solution to the system of equations Y= 2/3*x + 3 X= -2

Mathematics

2 answers:

Sergeu [11.5K]2 years ago

7 0

Answer:

( - 2, 1 and 2 / 3 )

Step-by-step explanation:

Check the procedure below;

$y = 2 / 3 * x + 3,\\x = - 2\\\\y = 2 / 3 * ( - 2 ) + 3,\\y = - 4 / 3 + 3,\\y = 5 / 3,\\x = - 2\\\\Conclusion; y = 5 / 3, and, x = - 2$

To derive this solution, we substituted the known value of x into the top equation, to receive the y value;

Solution - ( - 2, 5 / 3 )

schepotkina [342]2 years ago

6 0

Answer:

y=-4/3+3 you would multiply the numerator by X

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A student wants to find point C on the directed line segment from A to B on a number line such that the segment is partitioned i

olganol [36]

Answer:

Point C at (-18/7)

Step-by-step explanation:

Point A is at -6 and point B is at 2.

So, the distance between A and B = 2 - (-6) = 8

point C on the directed line segment from A to B on a number line such that the segment is partitioned in a ratio of 3:4.

Let the distance AC = x

∴ BC = 8 - x

AC : CB = 3 : 4

∴ $\frac{AC}{CB} =\frac{3}{4} = \frac{x}{8-x}$

Using cross multiplication

3 (8-x) = 4x

24 - 3x = 4x

24 = 7x

x = 24/7

So, Point C = -6 + 24/7 = -18/7

Point C at (-18/7)

7 0

3 years ago

Read 2 more answers

Classify ABC by its sides. Then determine whether it is a right triangle.

m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

$AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

AB = $\sqrt{(9-3)^{2}+(6-3)^{2} } = \sqrt{6^{2}+3^{2} } =\sqrt{45}$

Given vertices are B(6,9) and C( 6,-3)

$B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(-3-9)^{2}+(6-6)^{2} } =\sqrt{12^{2} } = 12$

BC = 12

Given vertices are A(3,3) and C( 6,-3)

$AC = \sqrt{(6-3)^{2}+(-3-3)^{2} } = \sqrt{9+36} = \sqrt{45}$

AC² = AB²+BC²

45 = 45+144

45 ≠ 189

∴Given Δ ABC is not a right angle triangle

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