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swat32
3 years ago
7

6x – 2y = 10 2x + 3y = 51 Solving the first equation above for y gives: y = x – 5

Mathematics
2 answers:
Virty [35]3 years ago
6 0

Answer:

<h2>x =6</h2><h2>y =13</h2>

Step-by-step explanation:

This is  the method I am familiar with.

I Hope It helps :)

METHOD- 1 : Elimination\\6x - 2y=10------(1)\\2x+3y =51------(2)\\Multiply -eq-(1)- by -the-coefficient-of-x-in-equation (2)\\Multiply-eq-(2) -by -the-coefficient-of-x-in-equation (1)\\6x - 2y=10------(1) *2\\2x+3y =51------(2)*6\\\\12x-4y=20 ------(3)\\12x+18y=306 ------(4)\\Subtract -eq- (4)- from- eq -(3)\\-22y =-286\\\frac{-22y}{-22} =\frac{-286}{-22} \\y =13\\

Substitute- 13- for y -in-equation -(1)-or-(2)\\6x - 2y=10------(1)\\6x -2(13)=10\\6x -26=10\\6x =10+26\\6x =36\\\frac{6x}{6} =\frac{36}{6} \\x =6

77julia77 [94]3 years ago
6 0

Answer:

Correct answers is

Step-by-step explanation:

1. 3

2. B

3. 6

4. (6,13)

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Demonstrates the commutative property across multiplication for 2 × 10 × 7
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3 0
3 years ago
Find the vertices and foci of the hyperbola with equation quantity x plus 4 squared divided by 9 minus the quantity of y minus 5
My name is Ann [436]

Answer:

Vertices at (-7, 5) and (-1, 5).

Foci at (-9, 5) and (1,5).

Step-by-step explanation:

(x + 4)²/9 - (y - 5)²/16 = 1

The standard form for the equation of a hyperbola with centre (h, k) is

(x - h²)/a² - (y - k)²/b² = 1

Your hyperbola opens left/right, because it is of the form x - y.

Comparing terms, we find that

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In the general equation, the coordinates of the vertices are at (h ± a, k).

Thus, the vertices of your parabola are at (-7, 5) and (-1, 5).

The foci are at a distance c from the centre, with coordinates (h ± c, k), where c² = a² + b².

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The coordinates of the foci are (-9, 5) and (1, 5).

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CROSS MULTIPLY
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6 0
3 years ago
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