Could someone help me with Algebra?

Question

This is confusing me! Please Help!

Solve the following systems of equations algebraically.

a.
5x-y=5
3x-5y=-63

b.
a-b+3c=-8
2b-c=15
3a+2c=-7

Answer 1

a) This is known as a "two equations, two unknowns" system of equations. There are two methods to solve this. I'll show you both of them.

(1) Elimination.

For elimination, what you do is multiply the equations by a given number, and add (or subtract) the equations to eliminate one variable. You then solve for that remaining variable. I'll explain:

5x - y = 5
3x - 5y = -63

Suppose we wanted to eliminate the x variable. What we would do is multiply the first equation by 3, and the second equation by 5. The goal here is to get 15x in both equation. Doing so, we get

15x - 3y = 15
15x - 25y = -315

Now, subtract the equations

(15x - 3y) - (15x - 25y) = 15 - (-315)

It should be automatic that the 15x cancels each other out. That will leave you with

-3y + 25y = 15 + 315
22y = 330
y = 15

Now that you have y = 15, you can plug this into ANY of the two equations to get the remaining variable. I'll pick 5x - y = 5.
5x - 15 = 5, 5x = 20, x = 4

So x = 4, y = 15 is the solution to the system of equations.

(2) Now we'll solve the same system of equations by substitution. In substitution, we solve for one of the variables in terms of the other in one equation, and plug in that for the other equation. Explanation for the given question below.

5x - y = 5
3x - 5y = -63

I'm going to solve for y in the first equation.
5x - y = 5, -y = -5x + 5, y = 5x - 5

Now, plug this value of y into the second equation, 3x - 5y = -63
3x - 5(5x - 5) = -63
3x - 25x + 25 = -63
-22x + 25 = -63
-22x = -88
x = 4.

Now that we have x = 4, we can similarly do what we did for elimination, and plug this into any of the two equations. I won't show the steps here, but you should get y = 15.

(b) To solve a "three equations, three unknowns" system, what you want to do is express one variable in terms of the other two in ONE of the equations, and then plug this value in for the other two, leaving you with a two equation, two unknowns problem.

a - b + 3c = -8
2b - c = 15
3a + 2c = -7

This time, I will choose the second equation to solve for c.
2b - c = 15, -c = -2b + 15, c = 2b - 15

Now, I will plug in c = 2b - 15 for the other two equations.

a - b + 3c = -8

Plugging in c = 2b - 15, we get

a - b + 3c = -8
a - b + 3[2b - 15] = -8
a - b + 6b - 45 = -8
a + 5b = 37 [Result after the first substitution]

Plugging in c = 2b - 15 for

3a + 2c = -7, we get

3a + 2[2b - 15] = -7
3a + 4b - 30 = -7
3a + 4b = 23 [Result after second substitution]

So now we have THESE two equations, two unknowns:

a + 5b = 37
3a + 4b = 23

Let's use substitution to solve these:
a + 5b = 37 implies a = -5b + 37

3[-5b + 37] + 4b = 23
-15b + 111 + 4b = 23
-11b = -88
b = 8

Plugging b = 8 into the equation a + 5b = 37, we get
a + 5(8) = 37, a + 40 = 37, a = -3

So now we have a = -3 and b = 8. We can now easily get c by plugging BOTH these a and b values into any one of the three original equations. Let's choose 3a + 2c = -7

3(-3) + 2c = -7
-9 + 2c = -7
2c = 2
c = 1

Therefore, our solution is a = -3, b = 8, c = 1.

Answer 2

question a.

5x - y = 5 equat 1
3x - 5y = -63 equat 2

mult equation 1 by 5

25x - 5 y = 25
3x - 5y = -63

add together

28x - 10y = -48

change signs

- 25x + 5 y = - 25
3 x - 5 y = - 63

add together

-22 x = - 88
x = 4

substitute in equation

5 x - y = 5

20 - y = 5

- y = - 20 + 5

- y = - 15
y = 15
Proof: 5x - y = 5

20 - 15 = 5

3x - 5 y = - 63
(3 x 4) - (5 x 15) = -63

12 - 75 = - 63


Question b

a - b + 3c = - 8 equation 1
2b - c = 15 equation 2
3a +2c = 7 equation 3

Mult equation 1 by 3

3a - 3b +9c = 24 equation 4
-3a - 2c = -7

add together

-3b + 7c = 17 equation 5
2b - c = 15 equation 2

mult 2 by 7

-3b + 7c = 17
14b - 7c = 105

add together

11b = 88
b = 8

Substitute in equation 2

2b - c = 15

-c = 15 -16 = -1

c = 1

substitute in equation 1

a - b +3c = -8

a - 8 + 3 = - 8

a = 8 - 8 - 3 = -3

Therefore a = -3
b = 8
c = 1

Answer 3

I'll walk you through problem a.
Basically, you want to solve for the variables one by one. You want to get an equation that says x = somenumber, or y = somenumber.

Here's the trick. Set one of the equations so that it says x = (some expression) or y = (some expression). I don't really feel like messing around with fractions, so let's use the first equation. You could use the second equation if you want, you'll get the same answer.

Shoving y over to one side, we get y = 5x - 5.

Now we take this, and plug it into the other equation in place of the y. That turns it into
3x - 5(5x-5) = -63.
We can replace y with the first equation since that's what "equals" means, right?

From there, just simplify.
3x - 25x +25 = -63
-22x = -88
x = (-88/-22)
x = 4

Now, to get y. Plug x into either equation, and you'll get a y value. Making sure that both equations give you the same value of y is also a way to double check your work.

y = 5x - 5
y = 5(4) - 5
y = 15

5y = 3x + 63
5y = 3(4) + 63
5y = 75
y = 75/5
y = 15

The equations agree that y = 15.
x = 4, y = 15 are our answers.

Answer 4

a. Use substitution.

From the first equation, we can solve for y to get y = 5x-5

Substituting this into the second equation we get:
3x-5(5x-5)=-63
3x-25x+25=-63
-22x=-88
x=4

Plugging x=4 into y = 5x-5, we get y = 5(4)-5 = 15

Solution: x = 4, y = 15

b. Use substitution again.

From the second equation, we get b = (1/2)(c+15)
From the third equation, we get a = (1/3)(-2c-7)

Substituting both a and b into the first equation, we get:
(1/3)(-2c-7) - (1/2)(c+15) + 3c = -8
(-2/3)c-(7/3)-(1/2)c-(15/2)+3c = -8
(11/6)c-(59/6)=-8
11c-59=-48
11c=11
c=1

Then, a = (1/3)(-2(1)-7)=-3
And, b = (1/2)(1+15) = 8

So our solution is a = -3, b = 8, c = 1.

Answer 5

a.
5x-y=5 ---> 5x-5= y, so substitute 5x-5 for y in 2nd equation
3x-5y=-63
3x -5(5x-5) =-63
3x-25x +25 = -63
-22x = -88
x=4, so y = 5(4) -5 = 20-5 = 15

b.
a-b+3c=-8 <--- Eq 1
2b-c=15 <--- Eq 2
3a+2c=-7 <---- Eq 3
Multiply Eq 2 by 2 and add result to Eq 3 getting:
4b+3a = 23 <--- Eq 4
Now multiply Eq 3 by 3 and result to Eq 1 getting:
5b + a = 37 <--- Eq 5
Now multiply Eq 5 by -3 and add result to Eq 4, getting:
-11b = -88
b=8
Put b=8 into Eq 5 getting:
5*8 +a = 37
40 +a =37
a = -3
Now put a=-3 and b= 8 into Eq1 getting:
-3 +8 +3c =-8
5+3c= -8
3c= -13
c= -13/3

Answer 6

In the 1st,
25x - 5y = 25
-3x + 5y = 63
------------------
22x = 88
x = 4

5(4) - y = 5
20 - y = 5
y = 15

For the 2nd, call the equations (I), (II), and (III)
add (I) and 3(II), get (IV) a + 5b = 37
add 2(II) and (III), get (V) 3a + 4b = 23,
then add 3(IV) and -(V), get 11b = 88, so b = 8

Then plug 8 into (II), 16 - c = 15, so c = 1.
Then plug 1 into (III), 3a + 2 = -7, 3a = -9, so a = -3.

Answer 7

a. Take the 1st or 2nd equation and solve for "x" or "y"...let's take the 1st equation and solve for "y" > subtract 5x from both sides:

5x - 5x - y = - 5x + 5
-y = -5x + 5 ("y" can't be negative-change all signs):
y = 5x - 5

First: take 5x - 5 and replace it with "y" in the 2nd equation:

3x - 5(5x - 5) = - 63
3x - 25x + 25 = - 63
-22x + 25 = - 63

Sec: subtract 25 from both sides and divide both sides by -22:

-22x + 25 - 25 = -63 - 25
-22x = - 88
-22x/-22 = - 88/-22
x = 4

Third: take 4 and replace it with "x" in the 1st equation:

5(4) - y = 5
20 - y = 5
20 - 20 - y = 5 - 20
- y = -15
y = 15

(4, 15)

Answer 8

a.
5x-y=5
3x-5y=-63

First you multiply everything on the top by -5 to make it +5y.

Now it is..
-25x+5y=-25
3x-5y=-63

The -5y and the +5y cancel out.
Now you have..
-25x=-25
3x=-63

You add them..
-25x+3x=-22x and -25+-63=-88
Now -22x=-88

You find out what x=?
-88/-22=x
x=4

Substitute x in the original problem.
5(4)-y=5
20-y=5
-y=5-20
-y=-15
y=15

B.
Sorry i cant help you with the second question! =[

Answer 9

A.
5x-y=5
3x-5y=63

5x-5=y
3x-5(5x-5)=-63

5x-5=y
3x-25x+25=-63

5x-5=y
-22x=-25-63

5x-5=y
-22x=-88

5x-5=y
-x=-88/22

5x-5=y
-x=-4

5x-5=y
x=4

5*4-5=y
x=4

y=15
x=4

Answer 10

(x - y^5)^3 = x^3 - 3x^2y^5 + 3xy^10 - y^15