# equation formula

#### Nollataulu

Hi. You guys know how to calculate (formula) these ones? I'm studying by myself, so I don't get any teacher help.

(k-3) â€“ (5-3k) = 2(k+1)

should it continue like this... (answer is k = 5)

k - 3 - 5 - 3k = 2k + 2

x/2 â€“ (x-1)/5 = 2

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#### topsquark

Math Team
Hi. You guys know how to calculate (formula) these ones? I'm studying by myself, so I don't get any teacher help.

(k-3) â€“ (5-3k) = 2(k+1)

should it continue like this... (answer is k = 5)

k - 3 - 5 - 3k = 2k + 2

x/2 â€“ (x-1)/5 = 2

For the first one, note the plus sign in front of the 3k
k - 3 - 5 + 3k = 2k + 2

For the second, what's the common denominator?

-Dan

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1 person

#### Nollataulu

For the first one, note the plus sign in front of the 3k
k - 3 - 5 + 3k = 2k + 2

For the second, what's the common denominator?

-Dan
Thanks a lot!

Second one it's number 10 I guess?

"The price of the computer rose by 8% in the spring but fell by 8% in the autumn, after that omputer cost â‚¬ 794.88. What was the price of the computer at the beginning of that year?"

Without having that fell by 8% section formula would be:

x + 0,08x = 794,88
1,08 =794,88
x = 736

But when having first going up and then down 8% not sure what kind of formula I should have?

#### skeeter

Math Team
Thanks a lot!

Second one it's number 10 I guess?

"The price of the computer rose by 8% in the spring but fell by 8% in the autumn, after that omputer cost â‚¬ 794.88. What was the price of the computer at the beginning of that year?"

Without having that fell by 8% section formula would be:

x + 0,08x = 794,88
1,08 =794,88
x = 736

But when having first going up and then down 8% not sure what kind of formula I should have?
Yes, 10 is the common denominator, so multiply every term by 10 to clear the fractions ...

$5x-2(x-1)=20$

take it from here?

percentage problem ...

$0.92(1.08x)=794.88$

2 people

#### Nollataulu

Yes, 10 is the common denominator, so multiply every term by 10 to clear the fractions ...

$5x-2(x-1)=20$

take it from here?

percentage problem ...

$0.92(1.08x)=794.88$
Thanks a lot about that second one And I guess first one should go like this.

5x-2 (x-1) = 20
5x-2x=20-2
3x=18
x=6

#### Nollataulu

How about this one? Jack buys MathForum shares for 2000 euros. How many shares will Jack receive if MathForum's share price is â‚¬6,50 and the acquisition cost is â‚¬8 plus 0,18% of the value of the shares to be purchased?

6,50x - 8 - 0,0018x = 2000 (?)

I'm not sure how I should deal with that 0,18%.

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#### DarnItJimImAnEngineer

Not quite, rather it's
Cost of shares : 6,50x
+ Acquisition cost : 8,00 + (0,0018*6,50x)
= Money spent : 2000,00
$1,0018 \cdot 6,50 ~â‚¬ \cdot x + 8,00 ~â‚¬ = 2000,00 ~â‚¬$

2 people

#### Nollataulu

Not quite, rather it's
Cost of shares : 6,50x
+ Acquisition cost : 8,00 + (0,0018*6,50x)
= Money spent : 2000,00
$1,0018 \cdot 6,50 ~â‚¬ \cdot x + 8,00 ~â‚¬ = 2000,00 ~â‚¬$
Thanks a lot

Forum Staff