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Sphinxa [80]
3 years ago
5

Which of the following equations have exactly one solution?

Mathematics
1 answer:
lukranit [14]3 years ago
3 0

Answer: B) 6x - 6 = 15x + 15

Step-by-step explanation:

To find our answer, we need to first work out each equation.

A) 6x - 15 = 6x + 15

First, cancel out the equal terms.

6x - 6x - 15 = 15

-15 = 15

This statement is false.

B) 6x - 6 = 15x + 15

-9x = 21

x = -7/3

<u>This statement has one solution. </u>

C) 6x - 6 = 6x + 15

-6 = 15

This statement is false.

D) 6x + 15= 6x + 15

15 = 15

This statement has infine solutions.

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Solve simultaneous equations 3x-2y=17 +2x-2y=10
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(7;2)

Step-by-step explanation:

3x - 2y  = 17 \\  - 2y = 17 - 3x \\ 2x + 17 - 3x = 10 \\ 2x - 3x = 10 - 17 \\  - x =  - 7 \\ x = 7 \\  - 2y = 17 - 21 \\  - 2y =  - 4 \\ y = 2

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Which expression is equivalent to (0.5n−0.3)−(0.8n−0.9)?
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-0.3n + 0.6

Step-by-step explanation:

if we write out the entire expression we have:

0.5n - 0.3 - 0.8n + 0.9 (two negative signs create a positive sign)

then we add and subtract

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-0.3n + 0.6

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3 years ago
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Solve for x in the equation 2x^2+3x-7=x^2+5x+39
Shalnov [3]
Hey there, hope I can help!

\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}
2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
x^2-2x-46=0

Lets use the quadratic formula now
\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}
x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

Multiply the numbers 2 * 1 = 2
\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}

\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \  \left(-2\right)^2=2^2, 2^2 = 4

\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \  \sqrt{4+184} \ \textgreater \  \sqrt{188} \ \textgreater \  2 + \sqrt{188}
\frac{2+\sqrt{188}}{2} \ \textgreater \  Prime\;factorize\;188 \ \textgreater \  2^2\cdot \:47 \ \textgreater \  \sqrt{2^2\cdot \:47}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \  \sqrt{47}\sqrt{2^2}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \  \sqrt{2^2}=2 \ \textgreater \  2\sqrt{47} \ \textgreater \  \frac{2+2\sqrt{47}}{2}

Factor\;2+2\sqrt{47} \ \textgreater \  Rewrite\;as\;1\cdot \:2+2\sqrt{47}
\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \  2\left(1+\sqrt{47}\right) \ \textgreater \  \frac{2\left(1+\sqrt{47}\right)}{2}

\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \  1+\sqrt{47}

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  2-\sqrt{188} \ \textgreater \  \frac{2-\sqrt{188}}{2}

\sqrt{188} = 2\sqrt{47} \ \textgreater \  \frac{2-2\sqrt{47}}{2}

2-2\sqrt{47} \ \textgreater \  2\left(1-\sqrt{47}\right) \ \textgreater \  \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \  1-\sqrt{47}

Therefore our final solutions are
x=1+\sqrt{47},\:x=1-\sqrt{47}

Hope this helps!
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Answer:

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Step-by-step explanation:

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