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natima [27]
4 years ago
8

Solve by substitution. Show each step of your work. 2x + y = 7 3x + 5y = 14

Mathematics
1 answer:
Monica [59]4 years ago
8 0
Alright...simple...showing all steps.. ;)

You have the equations...

2x+y=7
and
3x+5y=14

To be able to even solve for any of the variables...multiply the equations by...2...and..3...

2x+y=7----*3--> 6x+3y=21
and
3x+5y=14-----*2--->6x+10y=28

Thus,

6x+3y=21
-
6x+10y=28

=========

-7y=-7

y=1

Now, plug y back into any of the original equations....we'll use 2x+y=7 in this case....

2x+(1)=7

2x+1=7
      -1  -1

2x=6

x=3

Thus, the point of intersection for these two equations is (3,1)
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\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x² + y² = 64

____________________________________

\large \tt Solution  \: :

Standard equation of circle is :

\qquad \tt \rightarrow \: (x - h) {}^{2}  + (y - k) {}^{2}  =  {r}^{2}

  • \textsf{h = x - coordinate of centre of circle}

  • \textsf{k = y - coordinate of centre of circle}

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Since the circle is centered at origin, h = k = 0

\qquad \tt \rightarrow \: (x - 0) {}^{2}  + (y - 0) {}^{2}  =  {8}^{2}

\qquad \tt \rightarrow \:  {x}^{2}  +  {y}^{2} = 64

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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