Questions on Algebra: Systems of equations that are not linear answered by real tutors!

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A two- digit is such that the sum of the digits is 11. When the number with the same digits reversed is subtracted from this number, the difference is 9. What is the number?
**
let u=units digit
let t=tens digit
u+t=11
t=11-u
original number=10t+u
reversed number=10u+t
10t+u-10u-t=9
10(11-u)+u-10u-(11-u)=9
110-10u+u-10u-11+u=9
18u=90
u=5
t=11-u=6
ans:
original number=65

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What is the solution to the following system of equations?
-8x - 6y = 52
8x + 0y = -16
-----------------
8x + 0y = -16
Solve for x
-----
Sub for x in the 1st eqn to find y

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6 x -6 y = -48 .............1
-10 x + 4 y = 92 .............2

equation1
when x= 0 y= 8 ( 0 , 8 )
when y = 0 x= -8 ( -8 , 0 )
Equation 2
when x= 0 y= 23 ( 0 , 23 )
when y = 0 x= -9.2 ( -9.2 , 0 )

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Where is the image

6x – 6y = -48
/6
x-y =8
y=x-8

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explain how to solve a quadratic equation by completing the square.
Using the example x^2 + 6x - 2 = 0
:
We complete the square by finding the third term, when we subtract 2 from both sides:
x^2 + 6x + ____ = 2
:
The third terms can be found by dividing the coefficient of x by 2 and squaring it.
In this case it is 6, half of 6 = 3, 3^2 = 9, so we add 9 to both sides
x^2 + 6x + 9 = 2 + 9
x^2 + 6x + 9 = 11
:
Factor to reveal a perfect square
(x+3)(x+3) = 11
or
(x+3)^2 = 11
:
Find the square root of both sides
x + 3 = +/-
x = -3 +/-
Two solutions
x = -3 +
and
x = -3 -

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Total value of the all coins = $1.65
Total number of coins = 15
Value of a 5 cents coin = 5/100 = $0.05
Value of a 20 cents coin = 20/100 = $0.2

Let
You have total number of 5 cents coins = x
You have total number of 20 cents coins = 15-x

Solution
(Value of a 5 cents coin)(number of 5 cents coins)(Value of a 20 cents coin)(number of 5 cents coins)=Total value of the all coins
(0.05)(x)+(0.2)(15-x)=1.65
0.05x+3-0.2x=1.65
0.05x-0.2x=1.65-3
-0.15x=-1.35
-0.15x/-0.15=-1.35/-0.15
x=9

You have total number of 5 cents coins = x = 9
You have total number of 20 cents coins = 15-x = 15-9 = 6



Check
======
Value of 5 cents coins = (0.05)(9)=0.45
Value of 20 cents coins = (0.2)(6)=1.2
---------------------------------------
Sum = 1.65
=======================================

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you are given that a point on the line is (0,3) and the slope is -3.
the slope intercept form of the equation for a straight line is:
y = mx + b
m is the slope and b is the y-intercept.
the point slope form of the equation for a straight line is:
y-y1 = m(x-x1)
x1,y1 is a point on the line.
m is the slope.
you can find the equation using either of these forms.
i will show you how to do it with both.

assume you start with the slope intercept form of the equation.
that form is:
y = mx + b
since the slope you are given is -3, then replace m with -3 to get:
y = -3x + b
you know that a point on the line is (0,3)
replace y with 3 and replace x with 0 to get:
3 = -3*0 + b which becomes:
3 = b
the y-intercept is 3.
replace b in the slope intercept form of the equation with 3 and you get:
y = -3x + 3

assume you start with the point slope form of the equation.
that form is:
y-y1 = m(x-x1)
m is the slope.
x1,y1 is a point on the line.
you know the slope is -3 so you replace m with -3 to get:
y-y1 = -3(x-x1)
you know that a point on the line is (0,3), so you set (x1,y1) = (0,3) and replace y1 in your equation with 3 and x1 in your equation with 0 to get:
y-3 = -3(x-0) which becomes:
y-3 = -3x
add 3 to both sides of this equation to get:
y = -3x + 3

you got the same equation either way as you should.
that's how you do it.

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Let K and d be the ages of Mr. King and his daughter, respectively. Then,

K = 4d

Four years ago, Mr. King was K-4 years old and his daughter was d-4 years old. Also, we have

K-4 = 6(d-4)

So now we have a system of two equations. We can replace K with 4d to obtain

4d-4 = 6(d-4) = 6d-24

You should get d = 10. Since K = 4d, K = 40.

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Two numbers, and

The sum:

One third of the sum:

One third of the sum is 12:

The difference:

Half of the difference:

...is 14:

Simplify the two equations:





Solve the 2X2 system. Hint: It is already set up for Elimination.

John

My calculator said it, I believe it, that settles it

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Solve it algebraically if the equation is easily solvable using algebraic methods. Use a graph or computer algebra system for equations that can't be solved algebraically, e.g.

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xy=3
y= 3/x
substitute y in the equation
5x-6y=6
5x -6*(3/x)=6
multiply the equation by x
5x^2-18=6x
5x^2-6x-18=0
Find the roots of the equation by quadratic formula

a= 5 , b= -6 , c= -18

b^2-4ac= 36 + 360
b^2-4ac= 396



x1=( 6 + 19.9 )/ 10
x1= 2.59
x2=( 6 -19.9 ) / 10
x2= -1.39

if x = 2.59 y = 3/2.59 = 1.16
(2.59,1.16)
If x= -1.39 y= 3/-1.39=-2.16
(-1.39,-2.16)

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{x+3y=6}
{4x + 6y = -6}
----------------
Multiply thru the 1st equation by 2:
2x + 6y = 12
4x + 6y = -6
----------------------
Subtract and solve for "x":
2x = -18
x = -9
---------------
Solve for "y":
x + 3y = 6
-9 + 3y = 6
3y = 15
y = 5
-------------------
Solution: (-9,5)
====================
Cheers,
Stan H.
These are the choices for the problem
(-9, 5)
(-3, 3)
(6, 3)
(12, -2)

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1. First, change the equations so that if one is either added or subtracted from the other, one of the variables (x or y) would cancel out. For example, you might change 2y = x - 4 to 4y = 2x - 8 so that when added to -4y = 3x + 7, the 4y and -4y would cancel out. In your equations, this is not necessary because when y and y are subtracted, they cancel one another out.
2. Subtract/add your equations. In your case, you would subtract so that the y's would cancel out.
y = 4x - 2
-(y = -2x + 4)
==> 0 = 6x - 6
3. Find the value of the non-canceled out variable.
0 = 6x - 6
+6 +6
6 = 6x
1 = x
4. Plug this value for x into one of the equations (it does not matter which one) and solve for y.
y = 4x - 2
y = 4(1) - 2
y = 4 - 2
y = 2
5. You now have your x and y values. Simply plug them into coordinate form (x,y) for your answer. ==> (1,2)

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Please help me solve this equation by elimination:


----------------------- Subtract

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x+2y+3z=6
y+2z=0
z=2
================================
Solutions:
z = 2
-------
So, y + 2*2 = 0 ; So, y = -4
-------
Then, x + 2*-4 + 3*2 = 6
x - 8 + 6 = 6
x = 8
====================
Cheers,
Stan H.

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is basically

By using FOIL
First-
=>
Outer-
=>
Inner-
=>
Last-
=>
Thus add all of the separate components together to get

Simplify

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Hi, let x be the number of months of operation. Let the functions f(x) and g(x) to be the costs of the first and second freezer respectively in terms of months owned. Then we have,
f(x)=1800+60x
g(x)=2600+40x

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a 4% salt solution is mixed with a 16% salt solution. How many milliliters of each solution are needed to obtain 600 milliliters of a 10% salt solution?
-------
10 is the average of 4 and 16, so it's equal amounts.

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Is this ?
In long form this is:

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What system?

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3 x -1 y = 11 .............1
1 x -3 y = 9 .............2
Eliminate y
multiply (1)by -3
Multiply (2) by 1
-9 x 3 y = -33
1 x -3 y = 9
Add the two equations
-8 x = -24
/ -8
x = 3
plug value of x in (1)
3 x + -1 y = 11
9 + -1 y = 11
-1 y = 11
-1 y = 2
y = -2

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a-2b=2.................(1)
a+b=2(b+3)
a+b=2b+6
a-b=6..................(2)

a-2b= 2
a-b= 6
Eliminate y
multiply (1)by -1
Multiply (2) by 2
-a+2b=-2
2a-2b=12
Add the two equations
a = 10

plug value of a in (1)
1 a + -2 b = 2
10 + -2 b = 2
-2 b = 2
-2 b = -8
b = 4.00

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The 3rd equation is garbled.
Cheers,
Stan H.

Lining the letters up vertically as below, makes it easier
to see what's what:

 3x + 3y      =   3
-3x      - 2z = -11
       y +  z =  -1

The idea is to reduce this 3x3 system to a 2x2 system.
This is done by using the three equations to obtain
only two equations in only two unknowns.  

1.  We observe that x is already eliminated from the 3rd
equation 

and 

2. We observe that x is easily eliminated from the 1st
and 2nd equations just by adding them term by term
just as they are, since the coefficients of x in them
are equal in absolute value but opposite in sign.
So we add them term by term:

 3x + 3y      =   3
-3x      - 2z = -11
-------------------
      3y - 2z =  -8
 
3.  Next we solve the system

3y - 2z = -8
 y +  z = -1

We can eliminate z if we multiply the 2nd equation (which
was the 3rd original equation) by 2 and adding them term by term:

3y - 2z = -8
2y + 2z = -2
------------
5y      = -10
      y = -2

4. Then we substitute -2 for y in

  y + z = -1
 -2 + z = -1
      z = 1

5. Then finally we substitute -2 for y in
the first original equation
 
   3x + 3y = 3
3x + 3(-2) = 3
    3x - 6 = 3
        3x = 9
         x = 3

Solution (x,y,z) = (3,-2,1)

Edwin