use the augmented matrix method to reduce the system of equations to triangular form.?

Question

Write the general solution to the system in the form (x(z), y(z), z)

2x + 3y + 4z = 1
3x + 4y + 5z = 3

Answer 1

[2 3 4 | 1]
[3 4 5 | 3]
~ (R2 -> 3R1 - 2R2)
[2 3 4 | 1]
[0 1 2 | -3]
This is now in triangular form, with corresponding equations
2x + 3y + 4z = 1
y + 2z = -3
So y = -3 - 2z and we get
2x + 3(-3 - 2z) + 4z = 1
=> 2x = 1 + 9 + 6z - 4z
=> x = 5 + z
So the solution is (z+5, -3-2z, z) for arbitrary z.

Answer 2

i visit imagine of a minimum of four ideas: a million) elimination (upload or subtract the equations to get rid of a variable) 2) substitution (rearrange one equation to get one equation on its own, then substitute into the different equation 3) Kramer's Rule (matrix operations) 4) Graphically (search for the intersection) definite, a gadget may have more effective than one answer-- if the strains are all an similar line, there are a limiteless form of thoughts (each and every aspect on the line is a answer). Algebraically, you eventually end up with some thing that appears like 0 = 0 or 3 = 3. A gadget may even haven't got any thoughts (if the strains are parallel, or if more effective than 2 strains intersect do no longer intersect in a unmarried aspect). Algebraically, you eventually end up with some thing that appears like 0 = 3, or 9 = 12. the way the equations are provided oftentimes ascertain it really is more convenient. ex: y = 3x - 5 2x + 3y = 18 ==> because the first equation is already solved for y, this can be more convenient to apply substitution (the 2d equation ought to change into : 2x + 3(3x - 5) = 18 4x - 5y = -8 2x + 5y = 26 ==> should be more convenient to remedy by elimination, in view that including the equtions ought to yield 6x = 18 note: each and every each and every now and then, if the first answer you calculate is a "messy fraction", it ought to then be more convenient to pass back and get rid of the different variable particularly than substitute the fraction back in to remedy for the different in spite of the indisputable fact that, finished outstanding, each and every technique will yield an similar answer. Graphically (by hand) is the trickiest to achieve on the right answer-- particularly even if it isn't an integer, or the coefficients aren't any more integers. A calculator seems after this.