#how do I get the value of the letters in algebra?
north obsidianBOT
woeful wadi
have you got an example question?
quartz kindle
if you have x - 3 = 7, you are looking for a number represented by the symbol x that satisfies the equation x - 3 = 7
You could say this in words. "A number that, if you took 3 away, you would get 7"
This number is, of course, 10. This can be found by just adding 3 and 7.
To do this with algebra, you take your equation and you reverse the subtraction of the 3 by adding 3 to both sides.
x - 3 = 7
x - 3 + 3 = 7 + 3
x = 10
Sometimes people write numbers and letters next to each other. This is for multiplication
3x is three times x
-2x = 12. Divide both sides by -2 to get the answer.
-2x / -2 = 12 / -2
x = -6
You can have combinations of multiplication and addition
6x + 2 = 20. We are looking for a number that, when multiplied by 6 and then increased by 2, yields 20. First subtract 2 from both sides.
6x + 2 - 2 = 20 - 2
6x = 18. Then we divide both sides by 6
6x / 6 = 18 / 6
x = 3
You can have variables on both sides
x + 1 = 3x - 5
To solve an equation like this, the strategy is to move all the variables to one side and the constants (a term without variables) to the other
Subtract x from both sides and add 5 to both sides
x + 1 - x + 5 = 3x - 5 - x + 5
6 = 2x
6 / 2 = 2x / 2
3 = x
By reflexive property of equality, we can flip this and rewrite it as x = 3. Usually we write the variable we found on the left side of an equation.
Note that you might also have divisions in the equations
x/4 - 2 = 1
x/4 - 2 + 2 = 1 + 2
x/4 = 3
4 * x/4 = 4 * 3
x = 12
We can also have exponents. If you can't recall, an exponent is when you repeatedly multiply a number. 2 ^ 4 = 2 * 2 * 2 * 2 = 16 and 6 ^ 3 = 6 * 6 * 6 = 216
You, of course, will have variables to the power of a number. This is where you use roots.
x^2 = 9
x = sqrt(9) (sqrt is square root)
x = 3 or x = -3
Notice this equation, x^2 = 9, has two solutions.
3^2 = 9 and (-3)^2 = 9 (when we have a negative to an exponent we put it in parentheses)
To clarify, if x is a number such that x^2 = 9, then we have either x = 3 or x = -3, but we don't say "and". We obtain two possible answers, but we don't say x = 3 and x = -3, we say x = 3 or x = -3.
Another way to solve this equation, x^2 = 9, is by factoring. When we solve an equation by factoring, the strategy is to move all the terms to one side and obtain an equation where we have some expression equal to 0.
So, we subtract 9 from both sides.
x^2 = 9
x^2 - 9 = 9 - 9
x^2 - 9 = 0
The expression x^2 - 9 can be rewritten as x^2 - 3^2. This is called a difference of squares. The formula for facotring differences of squares is a^2 - b^2 = (a - b) (a + b)
So, that means x^2 - 9 = (x - 3) (x + 3)
We have that (x - 3) (x + 3) = 0
This is interesting. We have something multiplied by something else equals to 0. Remember that 0 times something is 0? And the product of two nonzero numbers is nonzero. This means that at least one of the factors is 0.
We write that x - 3 = 0 OR (here it is again) x + 3 = 0
We must solve both equations.
x - 3 = 0. Add 3 to both sides
x - 3 + 3 = 0 + 3
x = 3
nowwe consider x + 3 = 0
x + 3 - 3 = 0 - 3
x = -3
So this is the same solution. x = 3 or x = -3.
Sometimes we'll have an equation that we can get into the from ax^2 + bx + c = 0 (such as x^2 - 4x = -3, by adding 3 to both sides)
To solve this equation we have multiple options:
factoring
completing the square
the quadratic formula
To factor an expression that looks like ax^2 + bx + c = 0, we first find two numbers that multiply to ac and add to b
Let's go back to x^2 - 4x = -3.
Adding 3 to both sides yields x^2 - 4x + 3 = 0
a = 1, b = -4, and c = 3
These numbers are called coefficients
ac = 1 * 3 = 3
and b = -4
We need to find 2 numbers that multiply to 3 and add to -4
These two numbers are -1 and -3. -1 * -3 = 3 and -1 + -3 = -4
What we do with these two numbers is split the b term (recall they add to b)
x^2 - 4x + 3 = 0
x^2 - 3x - 1x + 3 = 0
Note that normally you will write 1x as x. This was just for emphasis on the splitting up of the term.
What you do with this is you group the left two terms and the right two terms.
(x^2 - 3x) + (-1x + 3) = 0
Now we factor each individually.
x^2 can be rewritten as xx.
x^2 - 3x = xx - 3x both have an x. We factor this out to write x^2 - 3x = x (x - 3).
Note the distributive property of muliplication a(b + c) = ab + ac
To factor out -1x + 3, we just take out the -1.
-1x + 3 = -1 (x - 3)
The magic is that both of these factored terms have x - 3, which we factor out
(x^2 - 3x) + (-1x + 3) = 0
x(x-3) + -1(x-3) = 0
(x-1)(x-3) = 0
This is great, because we just discussed how to solve something that looks like this.
x - 1 = 0 or x - 3 = 0
These two equations give us the two solutions x = 1 or x = 3
We can check these solutions back into our original equation, x^2 - 4x = -3
Is it true that 1^2 - 4(1) = -3? Yes, because on the left hand side you get 1 - 4
Is it true that 3^2 - 4(3) = -3? Yes, becuse on the left hand side you get 9 - 12
So that was factoring for quadratics.
Now we can move on to completing the square
When an equation is of the form x^2 + bx + c = 0, (if the coefficient of x^2 isn't 1, then you can divide everything by that coefficient to get it in that form) we can complete the square
Completing the square is a little magical at first sight. What we do is we take the b term, divide it by two, and then square that result.
We move over the c to the right side to get it to look like x^2 + bx = -c and then we add that new term to both sides
x^2 + bx + b^2 / 4 = b^2 / 4 - c
We can get back to doing it with variables afterwards, let's try it with numbers.
x^2 + 6x + 8 = 0. For reference, if we were to factor this, we'd get (x + 2) (x + 4) = 0, which tells us x = -2 or x = -4.
The b term is 6.
Divide it by two you get 3
Square that (multiply it by itself) you get 9
Let's move over the 8
x^2 + 6x + 8 - 8 = 0 - 8
x^2 + 6x = -8
Now we add 9, the number we have just obtained, to both sides
x^2 + 6x + 9 = 9 - 8
x^2 + 6x + 9 = 1
What we do is we factor the left hand side.
What multiplies to 1 * 9 = 9 and what adds to 6?
3 and 3 because 3 * 3 = 9 and 3 + 3 = 6
So we split up the b term:
x^2 + 6x + 9
= x^2 + 3x + 3x + 9
= (xx + 3x) + (3x + 3*3)
= x(x+3) + 3(x+3)
= (x+3) (x+3)
= (x+3)^2
This is very useful. We have something squared equals a number. All we have to do is take the square root of both sides.
x^2 + 6x + 9 = 1
(x+3)^2 = 1
x + 3 = sqrt(1)
x + 3 = 1 or x + 3 = -1
Taking the first eqaution, we have
x + 3 = 1
x + 3 - 3 = 1 - 3
x = -2
Taking the second equation, we have
x + 3 = -1
x + 3 - 3 = -1 - 3
x = -4
So, overall, we get x = -2 or x = -4, the same answer we get using factoring.
Completing the square allows us to create a sort of formula for these quadratics. I won't do it here, but, if you'd like, you can attempt to prove the quadratic formula by following the steps of completing the squares but keeping all of the coefficients as variables.
+/- means "plus or minus"
It tells us that there are two solutions
x = -b + ... or x = -b - ...
b^2 - 4ac is called the discriminant.
Sometimes quadratic equations won't have real solutions, such as x^2 + 1 = 0. No real number, when squared and then increased by 1, will yield 0, because the square of any real number will be greater than or equal to 0.
We know a quadratic equation ax^2 + bx + c = 0 won't have any real solutions if b^2 - 4ac is less than 0
If it equals 0, than the quadratic equation has 1 solution
If it is greater than 0, then the quadratic equation will have two solutions.
So that's quadratic equations. Fun, aren't they? They aren't as intuitive as the usual linear equations like 2x - 3 = 5.
Sometimes you'll have TWO equations with TWO unknowns, like
x + y = 5
2x - y = 1
What this means is we have two numbers, x and y, where their sum is 5, and the difference between twice of x and y is 1
These also have an interesting trick to them.
Usually, you multiply both equations by something to get a certain variable to have the same or opposite coefficients in the equations.
In the example equation, since in the first equation y has coefficient 1 and in the second equation y has coefficient -1, they are opposite (one is negative of the other) so we don't need to do anything.
What we do is ADD the two equations
You would subtract them if the certain variable had the same (not opposite) coefficient.
(x + y) + (2x - y) = 5 + 1
Hopefully you see how that is adding the two equations
By communitive and associative property of addition, we group the terms based on what variable they have
(x + 2x) + (y - y) = 6. Recall x is the same as 1x, so we can write x + 2x = 1x + 2x = 3x
3x = 6
3x / 3 = 6 / 3
x = 2
Then, we can choose one of our equations and input the value for the variable we just got to create an equation that allows us to find the value of the other variable.
We can choose x + y = 5, because that is the simpler equation.
Putting in x = 2, we get 2 + y = 5
2 + y - 2 = 5 - 2
y = 3
That means our two numbers are x = 2 and y = 3. Note that this time we are using "and," and that these are different variables, not the same variable. You might write it as an ordered pair (x, y), so we have (2, 3) is our solution.
These are called systems of equtions
That's basically all of the equaitons you do in algebra 1.
In algebra 2, you will review all of this and then do more stuff with logarithems and exponentials and trigonometric functions.
You might have an equation that looks like
3^x = 81
This is pretty easy, just put in numbers until you get the answer.
x = 4 because 3^4 = 81
However, to do this with algebra, we use something called logarithms
Because 3 is the base in the expression 3^x, we use the function "log base 3"
You take log base 3 (you can write it in text as log_3, and on a paper you would write log and then 3 as a subscript) of both sides
log_3 (3^x) = log_3 (81)
x = log_3 (81)
In a calculator, to do log_b (a) you put in log(a)/log(b)
lone nacelle
Lol
quartz kindle
log is usually log base 10 and ln is log base e (which is a number that is approximately 2.7182818, it is extremely fascinating and is very prevalenti n calculus)
lone nacelle
Is it copy paste?
quartz kindle
Thanks.
quartz kindle
I imagine that "value of letters in algebra" means he will just need to know stuff like x+2 = 4 and 3x = 9
Because those are the equations you do when you areintroduced to algebra
log_3 (81) = 4, so you get x = 4. Just like square roots, when a logarithm doesn't return an integer we normally just use as calculator.
Numbers like log_2 (3) are irrational and can't be expressed as a ratio of two integers.
To clarify, log_b (a) is referring to the number such that b ^ (that number) = a
This is an approximatino of log_2 (3)
This means that 2 ^ 1.584962... = 3
Another way you might be asked to solve 3^x = 81 is just write 81 as 3^4 and then equate the exponents
3^x = 81
3^x = 3^4
x = 4
This might be useful for something like 2^(2x-1) = 512
2^(2x-1) = 2^9
2x - 1 = 9
2x - 1 + 1 = 9 + 1
2x = 10
2x / 2 = 10 / 2
x = 5
Something that is odd about what we are doing is that 2 ^ 1.584962... doesn't really make sense. 1.584962... is not an integer. How can you multiply 2 by itself 1.584962... times? The answer is that you don't. I think we should explain noninteger exponents before we continue.
There are "laws of exponents," which are equations about exponents that are (almost) always true.
x^(a+b) = x^a * x^b
x^(a - b) = x^a / x^b
(x^a)^b = x^(ab)
x^(-a) = 1 / x^a
x ^ 0 = 1
x ^ 1 = x
All of these equations make sense in the context of a and b being integers.
For example, x^(a+b) is x times itself a+b times, which, by associative property of multiplication, would just be x times itself a times multiplied by x times itself b times.
We can use these equations to extend the definition of exponentiation to the rational numbers.
A fun one is to the power of 1/2.
Let's try x = 81^(1/2). Whatever it is, we want it to be greater than 0 because that'd make sense.
notice that x^2 = (81^(1/2))^2, which, by one of the laws of exponents, we get x^2 = 81^1, which, by another law of exponents, is 81.
x^2 = 81. All we have to do is take the square root of both sides to find x.
x = +/- 9. Since x > 0, x doesn't equal -9. This means x = 9
So we have 81^(1/2) = 9
We can repeat this similarly with x^(1/2), and we find that that is equal to sqrt(x)
Continuing on, we define x^(1/q) to be the qth root of x
Raise it to the power of p to get x^(p/q)
We define x^(p/q) to be the qth root of x to the power of p
For example, to find 125^(4/3), we first find the cube root of 125. That is 5, because 5 * 5 * 5 is 125. Next, we find 5 to the power of 4. that is going to be 5 * 5 * 5 * 5, which is 625. 125^(4/3) = 625.
We use continuity, which is a concept you would learn in calculus, to define exponents for irrational numbers.
For 2^1.584962... we can take a sequence of rational numbers approaching our irrational number, such as 1, 1.5, 1.58, 1.584, etc.
Then, that generates a sequence of numbers 2^1, 2^1.5, 2^1.58, 2^1.584, etc.
This sequence, when you calculate each term, gets closer and closer to 3.
To understand the logarithm more in depth, you might want to learn functions and then inverse functions. log_b (x) is the inverse of b^x
But those aren't really solving equations.
Because we have laws of exponents, we also have logarithm rules.
log(ab) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log(a^b) = b * log(a)
log(1) = 0
You might be given an equation involving logarithms, such as log_2 (3x - 4) + log_2 (x) = 1
You would use these log rules to get it into the form log_2 (...) = 1
Then you would change it from the logarithm form to the exponential form to get ... = 2^1
I don't know if there are any solutions, because I just put random numbers that time.
But that's logarithms and exponentials.
equations with trigonometric functions (sin, cos, tan, csc, sec, cot) you would need to laern the unit circle.
Sometimes you might have more complicated equations:
x + 6/x = -5
x = 2sqrt(x-1)
x^ln(x) = 1
sharp coral
How tf you go from introduction to algebra to mf natural logs 💀
quartz kindle
I explained natural logs
To solve more complicated equations, it just takes practice and trial and error. You have to understand what form you need to get the equation in to do certain things.
You can probably find videos of solving tough equations online or find a sheet of practice problems with more complicated equations
I'll go over the one with the natural log because it is slightly interesting.
When you have an expression with a variable to the power of a variable, normally you want to do e to the power of ln of it.
x^ln(x) = 1
e^(ln(x^ln(x)) = 1 This is in the form to apply one of the logarithm rules listed above
e^(lnx * lnx) = 1. One of the exponent rules tells us that e^0 is 1, so we get
e^(lnx)^2 = e^0
(lnx)^2 = 0
ln(x) = 0
e^ln(x) = e^0
x = 1
woeful wadi
godamn dedicated