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1 year ago

Solve system of equation algebraically y = 2x - 3 x + y = 18

Mathematics

2 answers:

Troyanec [42]1 year ago

8 0

Answer:

X=3 Y=15

Step-by-step explanation: First you need to use the problem 18/2=x to find the value of x. Now to find the value of y, you need to subtract 3 from 18. Then you solve

15=2(9)-3 15+3=18

natima [27]1 year ago

3 0

Answer:

x = 7

y = 11

Step-by-step explanation:

Given the system;

y = 2x - 3

x + y = 18

1. Approach

The easiest way to solve this system of equations is to solve the second equation for the variable (y). Then add the systems, use algebra to solve for the value of (x), then substitute that value back into one of the original equations to solve for the value of (y). Another name for the method in use is the method of elimination, this is when a [erspm manipulates one of the equations in a system of the equation such that when they add the equations, one of the variables eliminatates. Thus, they can solve for the other variable and the backsolve for the value of the unknown variable.

2. Solve one of the equations for a variable

Manipulate the system such that each equation is solved for the same variable,

x + y = 18

Inverse operations,

x + y = 18

-18 -18

x + y - 18 = 0

-y -y

x - 18 = -y

3. Use elimination

Now substitute this back into the original system,

y = 2x - 3

-y = x - 18

Add the systems,

y = 2x - 3

-y = x - 18

_________

0 = 3x - 21

Inverse operations,

0 = 3x - 21

+21 +21

21 = 3x

/3 /3

7 = x

4. Find the value of the unknown variable

Backsovle to find the value of (y),

x + y = 18

Substitute,

7 + y = 18

Inverse operations,

7 + y = 18

-7 -7

y = 11

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2 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

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Answer:

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Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

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Hence the apparent brightness of Alpha Centauri A is

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According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

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Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

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