Algebra math help, how to Get it?

Question

please show me the Steps to get the answer ,not just the answer..I really need help on these probs.

x^2 + 5 = 30 and y = 3x - 1 with x = 5 - 2y

Thanks!

for the y= 3 x -1 with x = 5 - 2y ..it says to solve for x and y


and x^2 + 5 = 30 ..says to solve for x..

Answer 1

First, if you're going to ask TWO QUITE SEPARATE QUESTIONS --- SEPARATE THEM!

You need to write something like: "Please solve:

a.) x^2 + 5 = 30, for x; and

b.) the two equations

(i) y = 3x - 1 and

(ii) x = 5 - 2y,

for x and y."

There, it's not so hard to re-express incomprehensible mathematical gibberish so that it makes sense, is it?!

That way, you won't see some poor intending responder floundering, unable to understand or even remotely discern what question(s) you're asking!

'Nuff said. Your INTENDED questions are solved below:

a.) x^2 + 5 = 30 ; subtract 5 from both sides, then

x^2 = 25, and therefore x = +/- 5.

b.) y = 3 x - 1, ......(i) and

.... x = 5 - 2 y. ......(ii)

Insert x from eqn. (ii) into eqn. (i). Then:

y = 3 (5 - 2 y) - 1, that is y = 15 - 6 y - 1 = 14 - 6 y.

Therfore, adding 6 y to each side, 7 y = 14, hence y = 2.

Put y = 2 back into eqn. (ii), then x = 5 - 2 y = 5 -2*2 = 1.

So the solution is (x, y) = (1, 2).

Now (VERY important!) CHECK that the original equations are indeed satisfied! :

If x = 1 and y = 2:

In eqn (i), the LHS = y = 2, the RHS = 3 x - 1 = 3*1 - 1 = 2. Good!

In eqn (ii), the LHS = x = 1, the RHS = 5 - 2 y = 5 - 2*2 = 1. Good!

So the solution found CHECKS OUT. QED

Live long and prosper.

Answer 2

Here you have TWO statements. (say them outloud)

X ^2 + 5 = 30 and Y = 3x - 1

There is also a condition here:
x = 5 - 2y

So you have two equations where ONE variable is unknown
(We know that x = 5 - 2y now)

So we will need to chose the equation that will most easily allow us to plug variable x in. (In this case it is the equation with ONE variable)

Step 1. (X) ^2 + 5 = 30, where X = 5 - 2y
Place the equivalent of X into the X being flanked by parentheses as follows:

Step 2. (5 - 2y)^2 + 5 = 30 then we can re-arrange the function

Step 3. (5 - 2y) * (5 - 2y) = 30 this is a "distributed version of the notation listed above in step 2

Use FOIL (First, Outside, Inside, Last) to distribute the functions (in this case you need to multiply the integers)

Step 4. 25 -10y -10y + 4y^2 = 30 (this is what your work should look like when you distribute.....we will need to clean this up a bit to make the next steps easier for you so pay attention to where things go.

Step 5. 4y^2 - 20y +25 = 30 Now we will need to set things up such that this equation will need to be equal to ZERO. (in its QUADRATIC form)

Step 6. 4y^2 -20y -5 = 0 We subtracted 30 from both sides and ended up with ZERO as a result. Now the equation is in its QUADRATIC form and we may proceed.

Step 7. 4Y^2 - 20y - 5 = 0 we"ll assign letter keys for each a b c -------> a=4, b= -20, c= -5

The quadratic can be found here:
http://en.wikipedia.org/wiki/Quadratic_equation

There are two answers for X:
a positive and a negative (on a graph these are your parabola or curve X-intercepts or some teachers call them ZEROs) This is the point at which the graph curve will touch the X-axis. Some curves don't touch the X-axis so the result are imaginary numbers but we won't worry about that yet.

Substitute the letters for the number values (include the negative or positive values)

and work the quadratic equation until you arrive to a positive and negative value for X.

Once that X is achieved, recall the untouched equation in step 1:
(X) ^2 + 5 = 30

plug both values for X in.....the good answer for X is the one that cancels the equation. 0=0 End of problem.


Or

you can chose to process the other equation first

Step 1. X^2 + 5 = 30
- 5 -5

x^2 = 25
Sq root (X^2) = SQ root (25)

x = +/- 5

then we can plug this into
X = 5 - 2y
5 = 5 - 2y
-5 = -5
0 = -2y not a solution
so let's try -5

-5 = 5 - 2y
-5 = -5
-10 = -2y isolate the y by dividing both sides by -2
5 = y

Answer 3

Problem 1: Quadratic equations --
x^2 + 5 = 30
Subtract 5 from both sides so variables are on one side and numbers are on the other: x^2 = 30 - 5 =25
Solve for x: x^2 = 25 so x = ±5 (2 correct answers)
Check for +5 answer: 5 ^ 2 +5 = 25 + 5 = 30
Check for -5 answer: -5 ^ 2 +5 = 25 + 5 = 30

Problem 2: simultaneous equations -- x and y have the same value in both equations.
(1) y = 3x - 1
(2) x = 5 - 2y
This can be solved 2 ways:
(a) substitute y from equation 1 into equation 2 and solve for x, then solve for y by substituting x back into equation 1. OR

(b) substitute x from equation 2 into equation 1 and solve for y, then solve for x by substituting y back into equation 2.

using procedure a.
(2) x = 5 - 2 (3x - 1) = 5 - 6x +2
so x = 7 -6x next add 6x to both sides
x + 6x = 7 - 6x + 6x next collect terms
7x = 7 or x = 7 / 7 or x = 1 answer for x.
Now put 1 in equation 1 wherever it says x
y = 3 (1) - 1 or y = 3 -1 or y = 2 answer for y

So, x = 1 and y = 2
Check by substituting these values in both equations
(1) y = 3x-1 or 2 = 3 - 1 or 2=2 True
(2) x= 5 - 2y or 1 = 5 - 2 (2) or 1 = 5 - 4 or 1=1 True.

Now you try to get the same answers using procedure b

davec996

Answer 4

I assume these are two separate problems.

The first one is pretty easy. What you want to do is isolate your variable on one side and your constant(s) on the other. The first step is to get rid of the "+ 5" on the left side. You can subtract 5 from both sides. Now you have a simple equation with a variable squared on the left and a constant on the right. Take the square root of both sides and you get the answer. Don't forget to plug the answer back into the original question to make sure it works.

The second one is a pair of equations in two variables. Again, you want to come up with a result that has one variable on one side and all constants on the other. You have two choices here. You can substitute the "3x-1" in place of the "y" in the second equation (because y = 3x-1, see?), and then use your regular tools to solve for x in the second equation. Alternatively, you can substitute "5-2y" in place of the "x" in the first equation, solve that equation for y. In either case, once you know the value of one variable, the other pops out of the other equation.

Let's try the first substitution:
x = 5 - 2 (3x - 1)
or x = 5 - 6x + 2

Now you can add 6x to both sides, simplify the constant, and get a simple equation "ax = constant"; divide both sides by "a" (whatever that turns out to be) and you get the value of x. Then y = 3x-1. Again, don't forget to plug these values of x and y into the second equation to make sure the answers are right.

You are smart enough to do this! Don't let anybody tell you that you're not.

Answer 5

since x=5-2y, substitute it into x^2+5=30
you should get 4y^2-10y=0
thus,
y=0 or y=2.5
sub into the middle equation of y=3x-1 and we get:
x=1/3 or x=7/6

Answer 6

x² + 5 = 30
x² = 25
x = ± 5

y = 3.(5 - 2y) - 1
y = 15 - 6y - 1
7y = 14
y = 2
x = 5 - 4
x = 1
Soln. is x = 1, y = 2

The second question may also be done by using simultaneous equations:-
3x - y = 1 X by 2
x + 2y = 5

6x - 2y = 2
x + 2y = 5----ADD
7x = 7
x = 1
1 + 2y = 5
2y = 4
y = 2
Soln. is x = 1 , y = 2 as before.

Answer 7

Really simple. so first one: subtract 5 on both sides, and we get x^2 = 25. which means what number multiplied by itself will get u 25? it would be positive and negative 5. second one:u have two equations, two variables. Just substitute. so we get y=3(5-2y) - 1. Multiply it out and y=2 and x is equal to 1.

Answer 8

What are the directions??