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Slav-nsk [51]
2 years ago
7

Write an equation of the line through (6,-5) having slope 0. Give the answer in standard form.

Mathematics
1 answer:
allsm [11]2 years ago
8 0

0x + 1y = -5 is the standard form of equation

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

Given that we have to write the equation of line passing thorugh (6, -5) and has slope 0

<em><u>The equation of line in slope intercept form is given as:</u></em>

y = mx + c ---------- eqn 1

Where, "m" is the slope of line and "c" is the y - intercept

Given that, m = 0

Given point is (6, -5)

<em><u>To find the y - intercept:</u></em>

<em><u>Substitute (x, y) = (6, -5) and m = 0 in eqn 1</u></em>

-5 = 0(6) + c

c = -5

<em><u>Substitute c = -5 and m = 0 in eqn 1</u></em>

y = 0x + (-5)

y = -5

<em><u>The standard form of equation is given as:</u></em>

The standard form of an equation is Ax + By = C

In this kind of equation, x and y are variables and A, B, and C are integers

y = -5

0x + 1y = -5 is the standard form of equation

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X-y+z=-4<br><br> 3x+2y-z=5<br><br> -2x+3y-z=15<br><br> How do I solve this?
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1)\ \ \ x-y+z=-4\ \ \ \Rightarrow\ \ \ z=-4-x+y\\\\2)\ \ \ 3x+2y-z=5 \\.\ \ \ \  \Rightarrow\ \ 3x+2y-(-4-x+y)=5 \\.\ \ \ \  \Rightarrow\ \ 3x+2y+4+x-y=5 \\.\ \ \ \  \Rightarrow\ \ 4x+y=1\\\\3)\ \ \ -2x+3y-z=15\\.\ \ \ \  \Rightarrow\ \ -2x+3y-(-4-x+y)=15\\.\ \ \ \  \Rightarrow\ \ -2x+3y+4+x-y=15\\.\ \ \ \  \Rightarrow\ \ -x+2y=11\\--------------------\\

z=-4-x+y\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\\ \left \{ {{4x+y=1\ \ \ \ } \atop {-x+2y=11\ /\cdot4}} \right. \\\\\left \{ {{4x+y=1\ \ \ \ } \atop {-4x+8y=44}} \right. \\-------\\y+8y=1+44\\9y=45\ /:9\\y=5\\\\-x+2y=11\ \ \ \Rightarrow\ \ \ x=2y-11\ \ \ \Rightarrow\ \ \ x=2\cdot5-11=-1\\\\z=-4-x+y=-4-(-1)+5=-4+1+5=2\\\\Ans.\ x=-1\ \ \ and\ \ \ y=5\ \ \ and\ \ \ z=2
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