Algebraically solve the system of equations using any method you wish
3x-10y+z=5
-4x+9y-z=-6
-13y+z=2
I would use substitution to start:
from the 3rd: z = 13y + 2
into the 1st:
3x - 10y + 13y + 2 = 5
3x + 3y = 3
x+y=1
into the 2nd:
-4x + 9y - 13y - 2 = -6
-4x - 4y = -4
x + y = 1
Ahhh,
x+y = 1 and x+y = 1 don't have a real solution.
So the system has no unique solution.
check my arithmetic
To algebraically solve the system of equations, we can use the method of substitution. Here's how:
Step 1: Solve one of the equations for one variable in terms of the others.
Let's solve the third equation for z:
-13y + z = 2
We can isolate z by adding 13y to both sides:
z = 13y + 2
Step 2: Substitute the expression you found in Step 1 into the other two equations.
Substitute z = 13y + 2 into the first equation:
3x - 10y + (13y + 2) = 5
Simplify:
3x + 3y + 2 = 5
Substitute z = 13y + 2 into the second equation:
-4x + 9y - (13y + 2) = -6
Simplify:
-4x - 4y - 2 = -6
Step 3: Solve the resulting system of two equations with two variables using another method, such as substitution or elimination.
Let's use the method of substitution again to solve this new system of equations.
From the equation 3x + 3y + 2 = 5, we can isolate x by subtracting 3y and 2 from both sides:
3x = -3y + 3
x = (-3y + 3) / 3
x = -y + 1
Now, substitute this value of x into the second equation:
-4(-y + 1) - 4y - 2 = -6
Simplify:
4y - 4 - 4y - 2 = -6
-6 = -6
Since -6 = -6 is a true statement, this equation is true regardless of the value of y. Therefore, y can take any value.
Step 4: Substitute the value of y back into one of the original equations to find the corresponding values of x and z.
Let's substitute y = a (where a represents any real number) into the first equation:
3x - 10a + z = 5
We can re-arrange this equation to solve for x:
3x = 10a - z
x = (10a - z) / 3
Now, we have the values of x, y, and z in terms of a:
x = (10a - z) / 3
y = a
z = 13a + 2
So the solution to the system of equations is:
x = (10a - z) / 3
y = a
z = 13a + 2
Note: This is the general solution since y can take any real value. To find specific solutions, substitute specific values for y.