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storchak [24]
3 years ago
14

Use substitution to solve the linear system of equations.

Mathematics
1 answer:
Ne4ueva [31]3 years ago
6 0

Answer:

Step-by-step explanation: infinitely many solutions (please give brainliest)

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A classmate simplified a rational expression below
lara31 [8.8K]
<h2>Part a) </h2><h2>Explain the error in this simplification.</h2>

Given the simplified expression

1-\frac{2}{x-2}=\frac{x+1}{x+2}

1-2\left(x+2\right)=\left(x+1\right)\left(x-2\right)

\:1-2x-4=x^2-x-2

-2x-3=x^2-x-2\:\:

0=x^2-2x-2\:\:

0=\left(x-1\right)\left(x-1\right)

x=1

<u><em>Identifying the Main Error</em></u>

1-\frac{2}{x-2}=\frac{x+1}{x+2}

1-2\left(x+2\right)=\left(x+1\right)\left(x-2\right)   ← ERROR Starts here

<u><em>Here is the Explanation of the Error </em></u>

<u><em /></u>

<u><em /></u>\mathrm{The\:equation\:should\:have\:been\:Multiplied\:by\:LCM=}\left(x-2\right)\left(x+2\right)<em>. </em>In your case you wrongly multiply the equation.

<em><u>CORRECTION</u></em>

<em>HERE IS HOW YOU SHOULD HAVE MULTIPLIED BY </em><em>LCM = (x-2)(x+2):</em>

<em />

1-\frac{2}{x-2}=\frac{x+1}{x+2}

1\cdot \left(x-2\right)\left(x+2\right)-\frac{2}{x-2}\left(x-2\right)\left(x+2\right)=\frac{x+1}{x+2}\left(x-2\right)\left(x+2\right)

\left(x-2\right)\left(x+2\right)-2\left(x+2\right)=\left(x+1\right)\left(x-2\right)

<h2>Part b) </h2><h2>Show your work as you correct the error</h2>

Here is the complete correction of the error.

Considering the expression

1-\frac{2}{x-2}=\frac{x+1}{x+2}

\mathrm{Find\:Least\:Common\:Multiplier\:of\:}x-2,\:x+2:\quad \left(x-2\right)\left(x+2\right)

1\cdot \left(x-2\right)\left(x+2\right)-\frac{2}{x-2}\left(x-2\right)\left(x+2\right)=\frac{x+1}{x+2}\left(x-2\right)\left(x+2\right)

\left(x-2\right)\left(x+2\right)-2\left(x+2\right)=\left(x+1\right)\left(x-2\right)

x^2-2x-8=x^2-x-2

x^2-2x-8+8=x^2-x-2+8

x^2-2x=x^2-x+6

-x=6

\frac{-x}{-1}=\frac{6}{-1}

x=-6

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Step-by-step explanation:

<u>I</u><u>d</u><u>e</u><u>n</u><u>t</u><u>i</u><u>t</u><u>y</u><u> </u><u>u</u><u>s</u><u>e</u><u>d</u><u> </u> :

\bold{ ({y}^{2} +   {z}^{2})  =  {y}^{2}  + 2yz +  {z}^{2}  }

\implies \:  {(2y)}^{2}  - 2.2y.3z +  {(3z)}^{2}

so we got :

y = 2

z = 3

{ \pink{ \bold{ \boxed{answer = (2y + 3z) {}^{2} }}}}

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