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NISA [10]
3 years ago
8

What value makes the equation 11-3x=-7 true?

Mathematics
2 answers:
german3 years ago
7 0
X = 6

11-3x=-7
Subtract 11 from both sides.

-3x = -18
Divide -3 on both sides.

X = 6
Virty [35]3 years ago
5 0

Answer:

11-3x=-7

11-3(6)=-7

11-18=-7

-7=-7

Step-by-step explanation:

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parantheses

Step-by-step explanation:

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3 years ago
What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S
Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is y=\frac{1}{3} x+2

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3  

And as the line passing through (-3,1) and is  perpendicular to 3x + y = -8, product of slopes of two line will be -1  as lies are perpendicular.

Let required slope = x  

\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

So we need to find the equation of a line whose slope is \frac{1}{3} and passing through (-3,1)

Equation of line passing through (x_1 , y_1) and having lope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

Hence the required equation of line is found using slope intercept form

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3 years ago
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3 years ago
7x + 2y = 24<br>8x + 2y = 30​
bagirrra123 [75]
(8x + 2y) - (7x + 2y) = 30 - 24

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