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3 years ago

2x + 3y= 22 -2x +5y=-6

Mathematics

1 answer:

sasho [114]3 years ago

5 0

The solution is x = 8 and y = 2

<em><u>Solution:</u></em>

<em><u>Given are the system of equations</u></em>

2x + 3y = 22 -------- eqn 1

-2x + 5y = -6 ----- eqn 2

We can solve the system of equations by elimination method

Eqn 1 + Eqn 2

2x + 3y -2x + 5y = 22 - 6

0x + 3y + 5y = 16

8y = 16

Divide both sides of equation by 8

Substitute y = 2 in eqn 1

2x + 3(2) = 22

2x = 22 - 6

2x = 16

Thus the solution is x = 8 and y = 2

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Solve the following system by any method -8x-8y=0 -8x+2y=-20

iVinArrow [24]

$\fbox{Solution by using Matrix}$

$\text{Linear Equation} = -8x-8y=0 , -8x+2y=-20$

$\text{Rewrite the linear equations above as a matrix}$

$\left[\begin{array}{ccc}8&-8&0\\-8&2&-20\\\end{array}\right]$

$\text{Apply to Row2 : Row2 + Row1}$

$\left[\begin{array}{ccc}8&-8&0\\-8&-6&-20\\\end{array}\right]$

$\text{ Simplify rows}$

$\left[\begin{array}{ccc}8&-8&0\\0&1&10/3\\\end{array}\right]$

$\text{Note: The matrix is now in echelon form.}$ $\text{The steps below are for back substitution.}$

$\text{Apply to Row1 : Row1 + 8 Row2}$

$\left[\begin{array}{ccc}8&0&80/30\\0&1&10/3\\\end{array}\right]$

$\text{ Simplify rows}$

$\left[\begin{array}{ccc}1&0&10/3\\0&1&10/3\\\end{array}\right]$

$\text{Therefore, the solution is}$

$x= \dfrac{10}{3} \ \text{and} \ \ y=\dfrac{10}{3}$

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3 years ago

HELP URGENT BRAINLIEST

nadya68 [22]

Given:

LMN is an equilateral triangle.

LM = LN = MN = 12 cm

To find:

The height of the triangle h.

Solution:

In a right angle triangle,

$\sin \theta=\dfrac{Opposite}{Hypotenuse}$

$\sin (60^\circ)=\dfrac{h}{12}$

$\dfrac{\sqrt{3}}{2}=\dfrac{h}{12}$

Multiply both sides by 12.

$\dfrac{\sqrt{3}}{2}\times 12=\dfrac{h}{12}\times 12$

$6\sqrt{3}=h$

Therefore, the height of the triangle is $6\sqrt{3}$ cm.

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tiny-mole [99]

Answer:

Its perpendicular

Step-by-step explanation:

You can solve the equation to get x=-2.5 and y = -7.5 so they intersect there, but they intersect at 90 degrees so its perpendicular.

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castortr0y [4]

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Andreyy89

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