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serg [7]
3 years ago
12

15-3yi=1/2x+2i solve for x and y

Mathematics
1 answer:
Aleksandr [31]3 years ago
3 0

Answer:

x=2(5−3iy−2i)

Step-by-step

Regroup terms.

5-3iy= \frac{1}{2} x+2i

​​Simplify \frac{1}{2}x\frac{x}{2}

5-3iy= \frac{x}{2}+2i

Subtract 2i from both sides.

5−3iy−2i= \frac{x}{2}

Multiply both sides by 2.

(5−3iy−2i)×2=x

Regroup terms

2(5−3iy−2i)=x

Switch sides

x=2(5−3iy−2i)

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OverLord2011 [107]
We will use the following law of indices (or 'index law') to check each pair of expression

x^{ \frac{m}{n}} = ( \sqrt[n]{x} )^{m}

With fractional power, the denominator is the root and the numerator is the power of the term. When the denominator is 2, we usually only write the normal square symbol (√). Denominator other than 2, we usually write the value of the root, for example, the cubic root ∛

Option A - Incorrect
7^{ \frac{5}{7}} should equal to ( \sqrt[7]{7} )^{5}

Option B - Correct
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7 0
3 years ago
Read 2 more answers
What is the equation of the line that passes through (-2,3) and is parallel to 2x+3y=6?
Vsevolod [243]

Answer:

<h2>2x + 3y = 5</h2>

Step-by-step explanation:

\bold{METHOD\ 1:}

The slope-intercept form of an equation of a line:

y=mx+b

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept</em>

Let k:y=m_1x+b_1,\ l:y=m_2x+b_2

then

l\ \parallel\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}

We have the equation of a line:

2x+3y=6

Convert to the slope-intercept form:

2x+3y=6           <em>subtract 2x from both sides</em>

3y=-2x+6              <em>divide both sides by 3</em>

y=-\dfrac{2}{3}x+2\to m_1=-\dfrac{2}{3}

therefore the slope is m_2=-\dfrac{2}{3}

Put the value of the slope and the coordinates of the point (-2, 3) to the equation of a line:

3=-\dfrac{2}{3}(-2)+b

3=\dfrac{4}{3}+b            <em>subtract 4/3 from both sides</em>

\dfrac{9}{3}-\dfrac{4}{3}=b\to b=\dfrac{5}{3}

Finally:

y=-\dfrac{2}{3}x+\dfrac{5}{3}

Convert to the standard form (Ax + By = C):

y=-\dfrac{2}{3}x+\dfrac{5}{3}              <em>multiply both sides by 3</em>

3y=-2x+5           <em>add 2x to both sides</em>

2x+3y=5

\bold{METHOD\ 2:}

Let k:A_1x+B_1y=C_1,\ l:A_2x+B_2y=C_2.

Lines <em>k</em> and <em>l</em> are parallel iff

A_1=A_2\ \wedge\ B_1=B_2\to\dfrac{A_2}{A_1}=\dfrac{B_2}{B_1}

We have the equation:

2x+3y=6\to A_1=2,\ B_1=3

then the equation of a line parallel to given lines has the equation:

2x+3y=C

Put the coordinates of the point (-2, 3) to the equation:

C=2(-2)+3(3)\\\\C=-4+9\\\\C=5

Finally:

2x+3y=5

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Answer:

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Step-by-step explanation:

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