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Sonbull [250]
3 years ago
12

Solve Systems by Substitution 2x+5y=-3 x+8y=4 and 2x+y=7 x-2y=-14

Mathematics
1 answer:
ASHA 777 [7]3 years ago
4 0
2x + 5y = -3 ⇒ 2x +   5y = -3 
1x + 8y =  4 ⇒ <u>2x + 16y = 8
</u>                               -<u>11y</u> = <u>-11 </u>
                                -11     -11
                                    y = 1
                        2x + 5(1) = -3
                            2x + 5 = -3
                            <u>      -5     -5</u>
                                  <u>2x</u> = <u>-8</u>
                                   2      2
                                    x = -4
                              (x, y) = (-4, 1)

2x + 1y = 7     ⇒ 2x + 1y = 7
1x -  2y = -14 ⇒ <u>2x  - 4y  = -28</u>
                                   <u>5y</u> = <u>35</u>
                                    5      5
                                     y = 7
                             2x + 7 = 7
                             <u>      -7   -7</u>
                                   <u>2x</u> = <u>0</u>
                                    2     2
                                    x = 0
                              (x, y) = (0, 7)
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The midpoint of segment QR is M(-2, 9). One end point of segment QR is Q(-8, 12). Find the coordinates of the other endpoint, R.
DanielleElmas [232]

Here,

The midpoint of segment QR is M

<u>m</u><u>(</u><u>-2</u><u>,</u><u>9</u><u>)</u> x3=-2,y3=9

<u>Q</u><u>=</u><u>(</u><u>-8</u><u>,</u><u>1</u><u>2</u><u>)</u>x1=-8,y1=12

R=x2,y2

we know that,

\tt{ m=\dfrac{Q+R}{2} } ⠀

So,

x3=\tt{\dfrac{x1+x2}{2} } ⠀

\tt{-2=\dfrac{-8+x2}{2}  } ⠀

\tt{-2×2=-8+x2  } ⠀

\tt{x2=8-4  } ⠀

\tt{x2=4  } ⠀

x2=4

now we find the y coordinate y2.

y3=\tt{\dfrac{y1+y2}{2} } ⠀

\tt{9=\dfrac{12+y2}{2}  } ⠀

\tt{9×2=12+y2  } ⠀

\tt{y2=18-12  } ⠀

\tt{y2=6  } ⠀

y2=6

<em>Now</em><em> </em><em>we</em><em> </em><em>get</em><em> </em><em>the</em><em> </em><em>two</em><em> </em><em>coordinate</em><em> </em><em>of</em><em> </em><em>R</em>

<em>so</em><em>,</em>

R=(x2,y2)=(4,6)

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