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Ganezh [65]
3 years ago
11

Write an equation perpendicular to x - 4y = 20 that passes through the point (4, -3)

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

y = - 4x + 13

Step-by-step explanation:

In order to do this problem you would need a whole new equation, but you will use what is given to you:

instead of going with:

x - 4y = 20

Turn it into y = mx + b form, also known as slope-intercept form:

x - 4y = 20

-x       = -x

_________

- 4y = 20 - x

Then divide both sides by - 4:

\frac{-4y}{-4} = \frac{-x + 20}{-4}

You will then get:

y = \frac{1}{4}x - 5

From this equation we will only need the slope!

m || = \frac{1}{4}

(This slope is for parallel)

but if you want perpendicular to the slope, then we need the negative reciprocal!

Negative reciprocal is a flipped version of the value that is negative, for example:

2 = -\frac{1}{2}

Because \frac{2}{1} = 2

Now we will find the perpendicular slope which is:

m ⊥ = - 4

Now substitute this slope into slope intercept form:

y = mx + b

(-3) = -4(4) + b

(Take away parentheses)

-3 = -16 + b

(Move - 3 to the other side, by making it positive, but what you do to one side, you do to the other)

= -13 + 6

(Move - 13 to the other side, by make - 13 into +13)

+13 = +13

________

13 = b

(This is you b, also known as you intercept)

Your answer is:

y = -4x + 13

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For what value of a should you solve the system of elimination?
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\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}

\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20
\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12

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6x + 10y = 20
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3a - 10y = -8

\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}

3a-10y=-8 \ \textgreater \  \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}
3a-10y-3a=-8-3a

\mathrm{Simplify} \ \textgreater \  -10y=-8-3a \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}-10
\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}

Simplify more.

\frac{-10y}{-10} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \  \frac{10y}{10}

\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \  y

-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{-8-3a}{-10}

\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \  -\frac{-3a-8}{10} \ \textgreater \  y=-\frac{-8-3a}{10}

\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \  6x+10\cdot \frac{8}{10-3a}=20

10\cdot \frac{8}{10-3a} \ \textgreater \  \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{8\cdot \:10}{10-3a}
\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \  \frac{80}{10-3a}

6x+\frac{80}{10-3a}=20 \ \textgreater \  \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}
6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}

\mathrm{Simplify} \ \textgreater \  6x=20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \  \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}

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\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}
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\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \  20\left(-3a+10-4\right) \ \textgreater \  Factor\;more

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\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \   \frac{10\left(-a+2\right)}{-3a+10}

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Hope this helps!
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