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Helga [31]
3 years ago
13

How many solutions does the system of linear equations 6a + 3b = 12 and 6a - 3b = 12 have?

Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

One solution

Step-by-step explanation:

Lets solve it for b

3b= -6a+12

3b= 6a-12

Simplifying

b= -2a+4

b= 2a+4

Both have same slope and intercept hence both lines are going to intersect at one point. So the equations are having one solution.

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AB=4x+3 and CD=24+5x find bc
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Answer:

Step-by-step explanation:

1. use distribution factor

(4x+3)(24+5x)

4x times 24 = 96

4x times 5x = 20x

3 times 24 = 72

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2. combine like terms

96 + 72 = 168

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35x + 168 or 168 + 35x

I may be wrong please check my answer. I may have a lapse in my algebra, or I may have misunderstood the problem. But this is what I can give you to the best of my knowledge.

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\bf 16sin^5(x)+2sin(x)=12sin^3(x)&#10;\\\\\\&#10;16sin^5(x)+2sin(x)-12sin^3(x)=0&#10;\\\\\\&#10;\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\&#10;-------------------------------\\\\&#10;2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=&#10;\begin{cases}&#10;0\\&#10;\pi \\&#10;2\pi &#10;\end{cases}\\\\&#10;-------------------------------

\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0&#10;\\\\\\&#10;~~~~~~~~~~~~\textit{quadratic formula}&#10;\\\\&#10;\begin{array}{lcccl}&#10;& 8 sin^4& -6 sin^2(x)& +1\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array} &#10;\qquad \qquad &#10;sin(x)= \cfrac{ -  b \pm \sqrt {  b^2 -4 a c}}{2 a}&#10;\\\\\\&#10;sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}&#10;\\\\\\&#10;sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=&#10;\begin{cases}&#10;\frac{1}{2}\\\\&#10;\frac{1}{4}&#10;\end{cases}

\bf \measuredangle x=&#10;\begin{cases}&#10;sin^{-1}\left( \frac{1}{2} \right)&#10;sin^{-1}\left( \frac{1}{4} \right)&#10;\end{cases}\implies \measuredangle x=&#10;\begin{cases}&#10;\frac{\pi }{6}~,~\frac{5\pi }{6}\\&#10;----------\\&#10;\approx~0.252680~radians\\&#10;\qquad or\\&#10;\approx~14.47751~de grees\\&#10;----------\\&#10;\pi -0.252680\\&#10;\approx 2.88891~radians\\&#10;\qquad or\\&#10;180-14.47751\\&#10;\approx 165.52249~de grees&#10;\end{cases}
3 0
3 years ago
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