How do I solve complex systems of equations with linear combinations?

Question

Apologies; Over the years I have forgotten.

When I ask "complex", I mean that you can't just multiply/divide 1 side to reach an equation that is simple to add/subtract to solve.

The system of equations I am currently attemping to solve is as follows:

3x - 4y = 21
4x + 2y = 6

I know how to solve this equation using substition and graphing but am unable to do it with linear combinations.

Thanks! Any assistance is greatly appreciated!

Answer 1

Any linear equations that can be put into the form

a0x+b0y+c0z+.......p0q=k0
a1x+b1y+c1z+.......p1q=k1
a2x+b2y+c2z+.......p2q=k2
*
*
anx+bnx+cnz+........pnq=kn

are called simultaneous' equations if they have (or we think that they may have) a common solution point.

Equations in this form can *always* be 'reduced' to the form (called 'upper triangular')

α0x+ß0y+σ0x+.......φ0p=µ0
0+ ß1y+σ1x+.......φ1p=µ0
0+ 0+ σ3x+.......φ3p=µ0
*
*
0+ 0+ 0+...........φnp=µn

by adding and/or subtracting multiples of various rows to other rows. When complete p=µn/φn and the rest of the variables may be found by continued back substitution.

This is called the 'Gauss-Jordan Elimination' method. If there is a problem and the family of equations 8cannot8 be put in upper triangular form, it means that there is no unique solution to the problem.

Consider the pair of equations in your question. Both of them describe straight lines (since both variables only appear to the 1'st power) and the 'solution' of this family is the {x,y} values that satisfy both equations. But what if the two lines described were parallel? Then there would be *no* solution. Or what if they were the same line? Then there would be an infinite set of solutions. (In your problem, neither case is true and there is a unique solution)

For your problem:

3x-4y=21
4x+2y=6 add twice the 2,nd equation to the 1'st

11x+0y=33 to get x = 3. Substitute into the 1'st to get
4y=9-21 = -12 to get y=-3

See how that works?


Doug

Answer 2

I really should get best answer for this:
Well there are four ways you can do it,
1. You can substitute in by solving for one variable, and then putting it into the second equations.
2. You can graph it and find the x,y intercept.
3. if you have a graphing calculator, you can use the simult equations solver, punch in each formula, and it will give you the values.
4. Or you can use a matrix to solve it. Like this:

Set up the intial equation like this.

3 -4 21
4 2 6


you need to do a series of math functions to change the equation to the matrix form

1 0 x
0 1 y

This will give you your answer.

so you start with
3 -4 21 (first divide the first row by 3)
4 2 6

1 -(4/3) 21 (now mult this row by -4 and add it to the bottom)
4 2 6


1 -(4/3) 21
0 (22/3) -22 (now divide this row by (22/3)

1 -(4/3) 21
0 1 -3 (now multi this row by -(4/3) and add it to the top)

1 0 3
0 1 -3

This is your answer x = 3, y = -3.

I

Answer 3

In a system of equations like
2x + 3y = 17
3x - 3y = 3,
it's easy to see that when adding both lines, the y's will drop out of the system, leaving 5x = 20 and a simple solution from there.

In your system, merely adding
3x - 4y = 21
4x + 2y = 6
doesn't eliminate either variable. What you'll have to do is to force the equations in such a way that a variable will drop out when adding them. The easiest way here is to multiply your bottom equation by two:
2(4x + 2y) = 2(6)
8x + 4y = 12
Now you'll have the system
3x - 4y = 21
8x + 4y = 12, which makes for simple elimination. 11x = 33, x = 3, and y = -3.

Here's one more example for you. Start with the system
4x + 7y = 6
6x + 5y = 20

Now what? Multiply some line through by a negative fraction? Yipes!
No, you can also multiply two lines, rather than just one. The x-coefficients are 4 and 6; their LCM is 12. You could multiply the top line by 3 (4 × 3 = 12) and the bottom by -2 (6 × -2 = -12) to have the x's drop out. If you'd rather solve for x and drop out the y's, the y-coefficients are 7 and 5; their LCM is 35. You could multiply your top line by -5 (7 × -5 = -35) and the bottom line by 7 (5 × 7 = 35). Either way makes for a valid solution... switching which line in either case you choose to be negative works, too. It's all a matter of what looks simplest to you.

Answer 4

What do you mean by:
"When I ask "complex", I mean that you can't just multiply/divide 1 side to reach an equation that is simple to add/subtract to solve."

Are you saying that you cannot find an multiple of 3 or -4 that is 4 and 2 respectively and vice-versa? Then you just created an imaginary wall to block your own way.

You can always multiply 1st eqn by 4/3 or 2/(-4) to so as to add/subtract to eliminate x or y, respectively, and solve for y or x. Similarly, if were to work on 2nd eqn, then multiply 2nd eqn by 3/4 or (-4)/2!

Answer 5

From eqn 2,
2y = 6 - 4x
or 4y = 12 - 8x...........multiplying both sides by 2.
Now, substituting above value of 4y i.e., 12 -8x in eqn 1
We get 3x - (12 -8x) = 21
i.e., 3x -12 + 8x = 21
or 11x = 21 +12
or 11x =33
i.e, x = 3
Substituting value of x i.e., 3 in eqn1 we get
9 - 4y = 21
or y = -3.
You can also solve this eqn using determinants. Hint:-
a1x + b1y +c1 = 0
a2x + b2y + c2 = 0
Thux/D1= Y/D2 = c/D3 where D =determinant and a,b and c are constants. Use correct sign conventions. Here c= 1

Answer 6

Put it in standad form:

The two equations come out to be y=-21/4+3/4X and y=3-2x in which -21/4 and 3 are the y-intercepts, and 3/4 and -2 are the slopes. I hope that helps.

Answer 7

Solve it like a matrix. I think someone here gave an example.

OR

A Ti-83 has the ability to solve like a matrix.
First make a 2x2 matrix (Matrix A), fill in
3 4
4 2
Then make a 1x2 matrix (matrix B)
21
6
Then go [A]^(-1)[B] : the first matrix to the power of -1, multiplied by matrix B.
You'll get your answers.,

Answer 8

With careful consideration and a calculator

Answer 9

its only your emagination...if you cant swim you look for a tuttor...not ans. or go back to k 1,2,3, that will be kindergarden..

Answer 10

go to school pal