How do i solve these equations algebraically?

Question

1) 5x-y=5
3x-5y=-63

2) how do i solve for all three variables
a-b+3c=-8
2b-c=15
3a+2c=-7

Also what are the two methosd of solvuing systems of equations algebracially?

Answer 1

The main methods are substitution and addition.

(1) To solve by addition, multiply one or both equations by a number so that the coefficients of one variable are additive inverses.

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

I'll multiply the first equation by -5 so that 5y will cancel out -5y in the second:

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

Now add them vertically:

-25x + 5y = -25
3x - 5y = -63
===============
-22x = -88
x = 4

Now I'll multiply the first by 3 and the second by -5 so that the x coefficients are 15 and -15 and cancel:

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

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

Now we can use either original equation to ensure that (4,15) is a solution:

5x - y = 5
5*4 - 15 =? 5
20 - 15 =? 5
5 = 5

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(2) To solve by substitution, place an equation in terms of one variable, and then substitute for that variable in another equation.

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

I'd place the second and third equations in terms of a and b:

2b - c = 15
2b = c + 15
b = (c + 15)/2

3a + 2c = -7
3a = -7 - 2c
a = -(7 + 2c)/3

Now substitute those into the first equation, replacing "a" with "-(7+2c)/3" and "b" with "(c+15)/2":

a - b + 3c = -8
-(7 + 2c)/3 - (c+15)/2 + 3c = -8

Multiply by 6 to get rid of the denominators and solve:

-(7 + 2c)/3 - (c+15)/2 + 3c = -8
-2(7 + 2c) - 3(c + 15) + 18c = -48
-14 - 4c - 3c - 45+ 18c = -48
11c = -48 + 14 + 45
11c = 11

c = 1

You already have equations for a and b in terms of c, so substitute c=1 in those equations to solve for a or b:

2b - c = 15
2b - 1 = 15
2b = 16
b = 8

3a + 2c = -7
3a + 2*1 = -7
3a = -9
a = -3

That's the solution: a=-3, b=8, c=1

Answer 2

Use linear algebra

1)
5x - y = 5, so y = 5x - 5
3x - 5y = -63, so 3x - 5(5x - 5) = -63
3x - 25x + 25 = -63
-22x + 25 = -63
-22x = -88
x = 4
y = 5(4) - 5 = 15

2)
Use linear algebra

a b c K
1 -1 3 -8
0 2 -1 15
3 0 2 -7

1 -1 3 -8
0 1 -.5 7.5
0 3 -7 17

1 0 2.5 -.5
0 1 -.5 7.5
0 0 -5.5 -5.5

1 0 2.5 -0.5
0 1 -.5 7.5
0 0 1 1

c = 1
b = 8
a = -3

Answer 3

question a million - upload 3 to the two components, so which you have 4x = 5. Divide the two components by means of four. you will get 5/4, although that's terrific to depart each and every thing in a fragment. question 2 - Subtract 10 from the two components, so which you have -a million = -x. Divide the two components by means of -a million to get a million = x. i wish this helps.

Answer 4

Why not use matrices? That way you use arithmetic and no algebra.