How do I solve for X in x/2 + x-3/5= 3

1 answer:

3 0

First of all,we have to cancel off the denominator.

times 2 to the equation in order to cancel off the denominator "2"
(x2) x+2x-6/5=6
#all numbers in this equation will be timed by 2 as well

then times 5 to cancel off the denominator"5"
(x5) 5x+10x-6=30
#all of them are timed by 5

then u will see this equation:
5x+10x-6=30

plus up the unknown numbers,
15x-6=30

Move the negative number "-6" to the back of 30.

When negative number moved to the other side,it will become positive number.
Same as when positive number moved to the other side,it will become negative number.

Thus,
"-6” moved
become positive "6"
Usually positive means plus"+" while negative means minus"-".

15x=30+6
15x=36

Move 15 to the opposite side,
Timed by 15 become divided by 15.

Just like the positive and negative numbers,they forced to become the opposite things.
positive→negative OR negative→positive
plus→minus OR minus→plus
times→divide OR divide→times

Thus,
x=36÷15
OR u can write as fraction,
x=36/15
x=12/5

final ans:
x=12/5 OR 2.4

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Write the quadratic equation in general form. (x - 3)^ 2 = 0

algol [13]

The answer is: x² – 6x + 9 = 0 .
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Explanation:
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Given: (x – 3)² = 0 ; write as: general form: "ax² + bx + c = 0"; a ≠ 0 .

Note: (x – 3)² = (x – 3)(x – 3) = x² – 3x – 3x + 9 = x² – 6x + 9 ;
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Rewrite: (x – 3)² = 0 ; →
as: x² – 6x + 9 = 0 ; which is our answer.
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→ x² – 6x + 9 = 0 ; is in "general form", or "standard equation format"; that is: " ax² + bx + c = 0 "; (a ≠ 0) ;
→ in which:
a = 1 (implied coefficient, since anything multiplied by "1" is that same value);
b = -6;
c = 9
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