The statement is false, as the system can have no solutions or infinite solutions.
<h3>
Is the statement true or false?</h3>
The statement says that a system of linear equations with 3 variables and 3 equations has one solution.
If the variables are x, y, and z, then the system can be written as:
Now, the statement is clearly false. Suppose that we have:
Then we have 3 parallel equations. Parallel equations never do intercept, then this system has no solutions.
Then there are systems of 3 variables with 3 equations where there are no solutions, so the statement is false.
If you want to learn more about systems of equations:
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They each give $10 because the pattern is changing by 10.
Move the decimal right 3 times
2.2512*10^8 let me know if its wrong
Answer:
Step-by-step explanation:
-2x + 3y – 4z = 8 ----------------(I)
5x – 3y + 5z = -8 -----------------(II)
7x – 3y + 3z = 8 ------------------(III)
Add equation (I) & (II) and thus y will be eliminated
(I) -2x + 3y – 4z = 8
(II) <u>5x – 3y + 5z = -8</u> {Add}
3x + z = 0 ------------------------(A)
Multiply equation (II) by (-1) and then add with equation (III). Thus y will be eliminated.
(II) * (-1) -5x + 3y - 5z = +8
<u>7x – 3y + 3z = 8</u> {Add}
2x -2z = 16 ---------------(B)
Multiply equation (A) by 2 and then add. Thus z will be eliminated and we will get the value of x
(A) * 2 6x + 2z = 0
(B) <u>2x - 2z = 16</u> {Add}
8x = 16
Divide both sides by 8
x = 16/8
x = 2
Plugin x = 2 in equation (A)
3x + z = 0
3*2 + z = 0
6 + z = 0
z = -6
Plug in x = 2 and z = - 6 in equation (I)
-2x +3y - 4z = 8
-2*2 + 3y - 4*(-6) = 8
-4 + 3y + 24 = 8
3y + 20 = 8
3y = 8 - 20
3y = -12
y = -12/3
y = -4
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