Find all solutions to the following system of equations:
x + 3y + 2z = 6
-3x + y + 5z = 29
-2x - 3y + z = 14
This just did not work out for me. Here is my work:
I first solved Equation #1 for x.
x = -3y - 2z + 6
Then, I did substitution, and replaced x for -3y - 2z + 6 in equations #2 and 3.
-3(-3y - 2z +6) + y + 5z = 29
-2(-3y - 2z +6) - 3y + z = 14
That is the same as:
10y + 11z = 47
and
3y + 5z = 26
Would elimination be the next step?
Thanks, Liese
Yes, elimination is the next step to solve the system of equations. To do elimination, multiply the second equation by 2 and the third equation by 3 to create a cancellation.
2(3y + 5z) = 2(26)
3(3y + 5z) = 3(26)
This simplifies the second and third equations to:
6y + 10z = 52
9y + 15z = 78
Now, subtract the second equation from the third equation to eliminate y:
(9y + 15z) - (6y + 10z) = 78 - 52
3y + 5z = 26
So, you have obtained the same equation as before:
3y + 5z = 26
This means that there is no cancellation and now you have two equations with the same variables, which means the system is dependent.
To solve the system, you can choose any expression for one variable and then solve for the other variables.
Let's solve for y:
3y + 5z = 26
Choose a value for y. Let's say y = t, where t is any real number.
Then:
3t + 5z = 26
We can rearrange this equation to solve for z:
5z = 26 - 3t
z = (26 - 3t) / 5
Now, we can substitute the values of y and z back into one of the original equations to solve for x. Let's use the first equation:
x + 3y + 2z = 6
Substitute y = t and z = (26 - 3t) / 5:
x + 3t + 2((26 - 3t) / 5) = 6
x + 3t + (52 - 6t) / 5 = 6
Multiply both sides by 5 to eliminate the denominator:
5x + 15t + 52 - 6t = 30
5x + 9t = -22
x = (-22 - 9t) / 5
So, the solutions to the system of equations are:
x = (-22 - 9t) / 5
y = t
z = (26 - 3t) / 5
Where t is any real number.
Yes, elimination is the next step to solve the system of equations. To use elimination, we want to eliminate one variable at a time by adding or subtracting the equations. In this case, we can eliminate either y or z.
Let's start by eliminating y. Multiply the second equation by 3 and the first equation by 1, then add them together:
(10y + 11z) + (3y + 5z) = 47 + 26
13y + 16z = 73
Now, we have a new equation:
13y + 16z = 73 ...(Equation 4)
Next, let's eliminate z. Multiply the third equation by 5 and the first equation by 2, then add them together:
(-2x - 3y + z) + (10x + 15y - 5z) = 14 + 70
8x + 12y - 4z = 84
Simplify the equation:
2x + 3y - z = 21 ...(Equation 5)
Now, we have two new equations: Equation 4 and Equation 5. To eliminate z, we can add Equation 4 and Equation 5:
(13y + 16z) + (2x + 3y - z) = 73 + 21
2x + 16y + 15z = 94
This gives us a third equation:
2x + 16y + 15z = 94 ...(Equation 6)
Now we have three equations: Equation 4, Equation 5, and Equation 6. To solve the system of equations, you can either continue using elimination or use substitution or matrix methods.
To find all solutions, you can solve this system of three equations and three variables using a method of your choice (such as elimination, substitution, or matrices). Once you find the values of x, y, and z, substitute them back into the original equations to make sure they satisfy all three equations.
I hope this helps! Let me know if you have any further questions.
why not use elimination right from the beginning
look at the y's
If you add the first and the last, they are gone
If you add 3times the second to the last, they are gone.
now you have 2 equations in x and z
not bad....
I finished it up like I said with elimination and got y = -3. Is this correct?
If not, I'll try it the way you suggested with elimination the whole way through.
Thanks, Liese