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Mathematics - algebriac equations

1. Solve 2x+5a/6 = 5x-3a for x in terms of a.

2. Make x the subject of y+6 = x+9/7x-1

The first question is actually 2x+ 5a = 5x-3a
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6

Ugh, 2x + 5a/6 = 5x-3a

$2x + \frac {5a}{6} = 5x-3a$

$\frac {5a}{6} + 3a = 5x - 2x$

$\frac {5a}{6} + \frac {18a}{6} = 3x$

$\frac {23a}{6} = 3x$

$\frac {23a}{18} = x$

2. Make x the subject of y+6 = x+9/7x-1, is the problem

$y+6 = x + \frac {9}{7}x-1$ (equation 2)

$y+7=\frac {16}{7}x$

$\frac {7}{16}(y+7)=x$

$\frac {7}{16}y + \frac {49}{16} = x$

That is, unless I misread your equation and it actually is something like $y+6=x+\frac {9}{(7x-1)}$ , in which case the equation is much more difficult.

Question 2 is actually like this:
y+6 = (x+9)/(7x-1)

Quote:

y+6 = (x+9)/(7x-1)

I assume you want x in terms of y. You almost have y in terms of x -- just subtract 6 from both sides.

$(y+6) = \frac {(x+9)}{(7x-1)}$

$(y+6)(7x-1)=(x+9)$

$42x + 7xy -y -6 = x+9$

$41x + 7xy = y+15$

$x(41+7y)=(y+15)$

$x = \frac {(y+15)}{(7y+41)}$

Thanks I understand how you did that one, except the question with the x-5<10x-23 I wouldn't have a clue how you got that answer.

$x-5<10x-23$ ?? I suspect you didn't follow exactly what I did. I put in too much. Copy it and compare it with this. I'll explain step-by-step what I did.

1. Add 23 to both sides.

$x+18<10x$

2. Subtract x from both sides

$18 < 9x$

3. Divide both sides by 9. Since 9 is a positive number the "sense" (<) of the inequality won't change. I forgot to do this step on your other thread.

$2 < x$

4. Don't like that? Turn it around.

$x \, > \,2$

Ohh I understand now, thanks so much.
And I have one more question if you don't mind:
Solve 4x^2+8x-1=0
I'm having trouble with these questions because I can't ask my teacher cause my year level has to stay home from school because of swine flu so I'm stuck with all this homework.

$4x^2+8x-1=0$

I'm not sure if you know the quadratic equation yet. This can be solved using that, or you can factor the equation

For a general quadratic equation of the form $ax^2 + bx + c=0$ the roots of the equation can be found from this equation:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ (quadratic equation)

using that we have

$x=\frac{-8\pm\sqrt{64+16}}{8}$

$x=\frac{-8\pm\sqrt{80}}{8}$

$x=-1\pm \frac{sqrt{5}}{2}$

The reason I did it that way is because of the irrational number. Usually in high school or middle school, they have problems that can be easily factored. You can actually see the factors using the roots of the first equation:

$(x-r_1)(x-r_2)$ where the r(s) are the roots of the equation. Here it is exactly:

$(x+(1+\frac {\sqrt 5}{2}))(x+(1-\frac {\sqrt 5}{2}))=0$

Here's an example of a typical problem:

$4x^2+7x-2=0$

$(4x-1)(x+2)=0$

If the product of two numbers equals zero, then one or both equals zero:

$4x-1=0\,\,or\,\,x+2=0$

Then x=1/4 or x=-2

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