Solving Linear Equations and Inequalities
Solve by eliminating x (this is solving 3 linear equations) then substitute to the other two equation
x + y + 5z =2 (1)
4x - 3y + 5z =3 (2)
3x - 2y + 5z=1 (3)
my teacher gave me the answer she said the solution set for this should be {(7,5-2)}
pleeeeeeeeeeeeeaaaseee :(
x = 2 - y - 5z
4(2 - y - 5z) -3y + 5z = 3
3(2 - y - 5z) -2y + 5z = 1
8 - 4y - 20z - 3y + 5z = 3
6 - 3y - 15z - 2y + 5z = 1
-7y - 15z = -5
-5y - 20z = -5 or y = 1 - 4z
-7(1-4z) - 15z = -5
7 + 28z -15 z = -5
13 z = 2
SO I get:
z = 2/13
y = 13/13 - 8/13 = 5/13
x = 26/13 - 5/13 -10/13 = 11/13
check
11/13 + 5/13 +10/13 = 26/13 = 2 yes
4(11/13)-3(5/13)+5(2/13)
=(44-15+10)/13 = 39/13 = 3
yes
33/13 -10/13 + 10/13 =33/13 = 1
yes
so I disagree
x + y + 5 z = 2
4 x - 3 y + 5 z = 3
3 x - 2 y + 5 z = 1
All three equations contain member 5 z
Equation ( 1 )
x + y + 5 z = 2 Subtract ( x + y ) to both sides
x + y + 5 z - ( x + y ) = 2 -( x + y )
5 z = 2 - x - y
Equation ( 2 )
4 x - 3 y + 5 z = 3 Subtract ( 4 x - 3 y ) to both sides
4 x - 3 y + 5 z - ( 4 x - 3 y ) = 3 - ( 4 x - 3 y )
5 z = 3 - 4 x - ( - 3 y )
5 z = 3 - 4 x + 3 y
Equation ( 3 )
3 x - 2 y + 5 z = 1 Subtract ( 3 x - 2 y ) to both sides
3 x - 2 y + 5 z - ( 3 x - 2 y ) = 1 - ( 3 x - 2 y )
5 z = 1 - 3 x - ( - 2 y )
5 z = 1 - 3 x + 2 y
5 z = 5 z = 5 z
2 - x - y = 3 - 4 x + 3 y = 1 - 3 x + 2 y Add x to all three sides
2 - x - y + x = 3 - 4 x + 3 y + x = 1 - 3 x + 2 y + x
2 - x + x - y = 3 - 4 x + x + 3 y = 1 - 3 x + x + 2 y
2 + 0 - y = 3 - 4 x + x + 3 y = 1 - 3 x + x + 2 y
2 - y = 3 - 3 x + 3 y = 1 - 2 x + 2 y Add y to both sides
2 - y + y = 3 - 3 x + 3 y + y = 1 - 2 x + 2 y + y
2 - 0 = 3 - 3 x + 4 y = 1 - 2 x + 3 y
2 = 3 - 3 x + 4 y = 1 - 2 x + 3 y Subtract 2 to all three sides
2 - 2 = 3 - 3 x + 4 y - 2 = 1 - 2 x + 3 y - 2
0 = 3 - 2 - 3 x + 4 y = 1 - 2 - 2 x + 3 y
0 = 1 - 3 x + 4 y = - 1 - 2 x + 3 y
Now you have system of 2 equations :
1 - 3 x + 4 y = 0
and
- 1 - 2 x + 3 y = 0
1 - 3 x + 4 y = 0 Add 3 x to both sides
1 - 3 x + 4 y + 3 x = 0 + 3 x
1 - 3 x + 3 x + 4 y + 3 = 0 + 3 x
1 + 0 + 4 y = 3 x
1 + 4 y = 3 x Subtract 1 to both sides
1 + 4 y - 1 = 3 x
1 - 1 + 4 y = 3 x - 1
0 + 4 y = 3 x - 1
4 y = 3 x - 1
- 1 - 2 x + 3 y = 0 Add 2 x to both sides
- 1 - 2 x + 3 y + 2 x = 0 + 2 x
- 1 - 2 x + 2 x + 3 y = 2 x
- 1 + 0 + 3 y = 2 x
- 1 + 3 y = 2 x Add 1 to both sides
- 1 + 3 y + 1 = 2 x + 1
- 1 + 1 + 3 y = 2 x + 1
0 + 3 y = 2 x + 1
3 y = 2 x + 1
You again have system of 2 equations :
4 y = 3 x - 1
and
3 y = 2 x + 1
4 y = 3 x - 1 Multiply both sides by 3
4 y * 3 = ( 3 x - 1 ) * 3
12 y = 3 x * 3 - 1 * 3
12 y = 9 x - 3
3 y = 2 x + 1 Multiply both sides by 4
3 y * 4 = 2 x * 4 + 1 * 4
12 y = 8 x + 4
12 y = 12 y
9 x - 3 = 8 x + 4 Subtract 8 x to both sides
9 x - 3 - 8 x = 8 x + 4 - 8 x
9 x - 8 x - 3 = 8 x - 8 x + 4
x - 3 = 0 + 4
x - 3 = 4 Add 3 to both sides
x - 3 + 3 = 4 + 3
x + 0 = 7
x = 7
Replace value of x in equation :
4 y = 3 x - 1
4 y = 3 * 7 - 1
4 y = 21 - 1
4 y = 20 Divide both sides by 4
4 y / 4 = 20 / 4
y = 5
Replace values x and y in equation :
5 z = 2 - x - y
5 z = 2 - 7 - 5
5 z = - 10 Divide both sides by 5
5 z / 5 = - 10 / 5
z = - 2
The soluitions are :
x = 7 , y = 5 , z = - 2
To solve the system of linear equations, we will use the method of elimination.
First, let's eliminate x from equations (2) and (3). We can do this by multiplying equation (1) by 4 and equation (3) by 3, and subtracting equation (1) from equation (2) and equation (3) from equation (2):
4(x + y + 5z) = 4(2) (Multiply equation (1) by 4)
3(3x - 2y + 5z) = 3(1) (Multiply equation (3) by 3)
This gives us:
4x + 4y + 20z = 8 (4)
9x - 6y + 15z = 3 (5)
Now we subtract equation (1) from equation (2) and equation (3) from equation (2):
(4x - 3y + 5z) - (x + y + 5z) = 3 - 2 (Subtract equation (1) from equation (2))
(9x - 6y + 15z) - (x + y + 5z) = 3 - 2 (Subtract equation (3) from equation (2))
Simplifying these equations, we get:
4x - 3y + 5z - x - y - 5z = 1
9x - 6y + 15z - x - y - 5z = 1
Combining like terms:
3x - 4y = 1 (6)
8x - 7y + 10z = 1 (7)
Now, we have a system of equations (6), (5), and (7).
Next, we will eliminate y by multiplying (6) by 9 and (5) by 3, and subtracting (5) from (6):
9(3x - 4y) = 9(1) (Multiply equation (6) by 9)
3(9x - 6y + 15z) = 3(3) (Multiply equation (5) by 3)
This gives us:
27x - 36y = 9 (8)
27x - 18y + 45z = 9 (9)
Subtracting equation (9) from equation (8):
(27x - 36y) - (27x - 18y + 45z) = 9 - 9
Simplifying this equation, we get:
-18y - 45z = 0 (10)
Now we have the system of equations (10), (5), and (7).
Solve equation (10) for y:
-18y = 45z
y = (45z) / (-18)
y = -5z/2 (11)
Now substitute equation (11) back into equations (5) and (7):
8x - 7(-5z/2) + 10z = 1
27x - 18(-5z/2) + 45z = 9
Simplifying these equations, we get:
8x + (35z/2) + 10z = 1
27x + 45z + (45z/2) = 9
Combining like terms:
8x + (35z/2) + 10z = 1 (12)
27x + (45z/2) + 45z = 9 (13)
Now, we have the system of equations (12) and (13).
To eliminate z, we can multiply equation (12) by 27 and equation (13) by 8, and subtract equation (12) from equation (13):
27(8x + (35z/2) + 10z) = 27(1) (Multiply equation (12) by 27)
8(27x + (45z/2) + 45z) = 8(9) (Multiply equation (13) by 8)
This gives us:
216x + 945z + 540z = 27 (14)
216x + 360z + 360z = 72 (15)
Subtracting equation (14) from equation (15):
(216x + 360z + 360z) - (216x + 945z + 540z) = 72 - 27
Simplifying this equation, we get:
-225z = -45
Dividing both sides by -225, we have:
z = -45 / -225
z = 1/5 (16)
Now substitute z = 1/5 back into equations (11) and (12):
y = -5(1/5) / 2
y = -1/2 (17)
8x + 35(1/5) + 10(1/5) = 1
Simplifying, we get:
8x + 7 + 2/5 = 1
Combining like terms, we have:
8x + 7/5 = 1
Subtracting 7/5 from both sides:
8x = 1 - 7/5
8x = 5/5 - 7/5
8x = -2/5
Dividing both sides by 8, we get:
x = (-2/5) / 8
x = -2/40
x = -1/20 (18)
Therefore, the solution set for the system of linear equations is {(x, y, z)} = {(-1/20, -1/2, 1/5)}.
To solve the system of linear equations given by equations (1), (2), and (3), you can use the method of elimination.
1. Start by eliminating x. To do this, we will perform operations on the equations to eliminate x.
Multiply equation (1) by 4:
4(x + y + 5z) = 4(2)
4x + 4y + 20z = 8
Multiply equation (2) by 3:
3(4x - 3y + 5z) = 3(3)
12x - 9y + 15z = 9
Multiply equation (3) by 3:
3(3x - 2y + 5z) = 3(1)
9x - 6y + 15z = 3
Now subtract the second equation from the first equation:
(4x + 4y + 20z) - (12x - 9y + 15z) = 8 - 9
-8x + 13y + 5z = -1 ........... (4)
Subtract the third equation from the first equation:
(4x + 4y + 20z) - (9x - 6y + 15z) = 8 - 3
-5x + 10y + 5z = 5 ........... (5)
2. Next, we will eliminate y. Multiply equation (4) by 10 and multiply equation (5) by 13:
10(-8x + 13y + 5z) = 10(-1)
-80x + 130y + 50z = -10
13(-5x + 10y + 5z) = 13(5)
-65x + 130y + 65z = 65
Now add the fifth equation to the fourth equation:
(-80x + 130y + 50z) + (-65x + 130y + 65z) = -10 + 65
-145x + 260y + 115z = 55 ........... (6)
3. Lastly, we will eliminate z. Multiply equation (4) by 23 and multiply equation (6) by 5:
23(-8x + 13y + 5z) = 23(-1)
-184x + 299y + 115z = -23
5(-145x + 260y + 115z) = 5(55)
-725x + 1300y + 575z = 275
Now add the sixth equation to the fourth equation:
(-184x + 299y + 115z) + (-725x + 1300y + 575z) = -23 + 275
-909x + 1599y + 690z = 252 ........... (7)
4. Now we have three equations (5), (6), and (7) with only y, z, and constant terms. This is a system of linear equations that can be solved by substitution.
Let's solve equations (5) and (6) for y and z:
From equation (5):
-5x + 10y + 5z = 5 ........... (5)
Rearranging the terms, we get:
10y + 5z = 5 + 5x
2y + z = 1 + x ........... (8)
From equation (6):
-145x + 260y + 115z = 55 ........... (6)
Rearranging the terms, we get:
260y + 115z = 145x + 55
52y + 23z = 29x + 11 ........... (9)
5. Now we substitute equations (8) and (9) into equation (7) to eliminate y and z:
-909x + 1599y + 690z = 252 ........... (7)
Substituting equations (8) and (9):
-909x + 1599(1 + x) + 690(29x + 11) = 252
Expanding and simplifying:
-909x + 1599 + 1599x + 29010x + 7590 = 252
18700x = -18099
x = -969/935
6. Substitute the value of x back into equations (8) and (9) to find the values of y and z:
From equation (8):
2y + z = 1 + (-969/935)
2y + z = -934/935
2y = -934/935 - z ........... (10)
From equation (9):
52y + 23z = 29(-969/935) + 11
52y + 23z = -28101/935 + 11
52y = -28101/935 + 10470/935 - 23z
52y = -17631/935 - 23z ........... (11)
7. Now we can choose a value for z and solve for y:
Let's set z = 1. Substituting into equations (10) and (11):
From equation (10):
2y + 1 = -934/935
2y = -934/935 - 1
2y = -1869/935
y = -1869/1870
From equation (11):
52y = -17631/935 - 23
52y = -18554/935
y = -18554/935 / 52
y = -2
8. Finally, substitute the values of x, y, and z back into one of the original equations (1) to check if the solution is correct:
x + y + 5z = 2
(-969/935) + (-2) + 5(1) = 2
-969/935 - 2 + 5 = 2
-969/935 - 2 + 5 = 10
Since the equation is true, we have found the solution to the system of equations. The solution set is {(x, y, z) = (-969/935, -2, 1)}.