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kirill115 [55]
2 years ago
14

11-1\2y=3+6x Solve for y

Mathematics
2 answers:
myrzilka [38]2 years ago
8 0

The answer is  y=4/3.                                                                                                                                

romanna [79]2 years ago
5 0
Let's solve for y.

<span><span>11−<span><span>12</span>y</span></span>=<span>3+<span>6x

</span></span></span>Step 1: Add -11 to both sides.

<span><span><span><span><span><span>−1/2</span></span>y</span>+11</span>+<span>−11</span></span>=<span><span><span>6x</span>+3</span>+<span>−11

</span></span></span><span><span><span><span>−1</span>2</span>y</span>=<span><span>6x</span>−8

</span></span>Step 2: Divide both sides by (-1)/2.

<span><span><span><span><span>−1/</span>2</span>y</span><span><span>−1/</span>2</span></span>=<span><span><span>6x</span>−8/-1/2

</span></span></span><span>y=<span><span>−<span>12x</span></span>+16</span></span><span>
</span>
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Find the positive base $b$ in which the equation $5_b \cdot 23_b = 151_b$ is valid.
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Answer:

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Step-by-step explanation:

We want to determine the positive base b in which:

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The easiest way to approach this is to convert all the numbers to base 10.

5_b=5Xb^0=5\\23_b =(2Xb^1)+(3Xb^0)=2b+3\\151_b=(1Xb^2)+(5Xb^1)+(1Xb^0)=b^2+5b+1\\Therefore:\\5_b \cdot 23_b = 151_b\\5(2b+3)=b^2+5b+1\\10b+15=b^2+5b+1\\b^2+5b+1-10b-15=0

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Next, we factorize the resulting expression.

b^2-5b-14=b^2-7b+2b-14=0\\b(b-7)+2(b-7)=0\\(b-7)(b+2)=0\\b-7=0\: or \: b+2=0\\b=7\: or \: b=-2

The positive value of b for which the equality hold is 7.

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