To solve this system of equations, we can use the method of elimination.
First, let's write down the system of equations again:
3x + 2y + z = 7
5x + 5y + 4z = 33
3x + 2y + 3z = 1
To eliminate a variable, we need to multiply the equations by appropriate coefficients so that when we add or subtract them, one of the variables will cancel out.
Let's start by eliminating z. We can multiply the first equation by 3 and the third equation by -1 to get:
9x + 6y + 3z = 7
-3x - 2y - 3z = -1
Adding these two equations together eliminates the z term:
9x + 6y + 3z + (-3x - 2y - 3z) = 7 + (-1)
This simplifies to:
6x + 4y = 6
Now let's eliminate z again. We can multiply the first equation by 4 and subtract it from the second equation:
5x + 5y + 4z - (12x + 8y + 4z) = 33 - (4 * 7)
Simplifying this equation gives:
-7x - 3y = -9
Now we have a system of two linear equations:
6x + 4y = 6
-7x - 3y = -9
We can solve this system using the method of substitution.
From the first equation, we can solve for x in terms of y:
6x = 6 - 4y
x = (6 - 4y) / 6
Now we substitute this expression for x into the second equation:
-7((6 - 4y) / 6) - 3y = -9
Simplifying this equation gives:
-42 + 28y - 18y = -54
Combining like terms gives:
10y = -12
Dividing both sides by 10 gives:
y = -12/10
y = -6/5
Now we substitute this value of y back into the expression for x:
x = (6 - 4(-6/5)) / 6
x = (6 + 24/5) / 6
Simplifying this expression gives:
x = (30/5 + 24/5) / 6
x = 54/5 / 6
x = (54/5) * (1/6)
x = 54/30
x = 9/5
Finally, we substitute these values of x and y back into any of the original equations to find z. Let's use the first equation:
3x + 2y + z = 7
3(9/5) + 2(-6/5) + z = 7
(27/5) - (12/5) + z = 7
(15/5) + z = 7
3 + z = 7
z = 7 - 3
z = 4
So the solution of the system of equations is x = 9/5, y = -6/5, and z = 4.
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What is the solution of the system of equations?
⎧⎩⎨⎪⎪3x+2y+z=75x+5y+4z=33x+2y+3z=1
1 answer