Solve each system.
x-4y-6z=-34
-7x+5y-7z=-24
z=5x-7y+18
These are my answers below but I know that they are wrong:
x=-3894
y=-112852
z=-107,509,144
I have tried to solve this system probably 10 times, and still not getting an answer that checks back into the original equation. Substitution doesn't work well and I do better with elimination.
5 answers
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
I get (4, 5, 3)
I get (4, 5, 3)
I'm sorry I don't really get matrices, but can you please walk me through the steps of algebraically doing elimination if you know what that is.
Can you do Gaussian elimination? http://www.purplemath.com/modules/systlin6.htm
or
http://hotmath.com/hotmath_help/topics/solving-systems-of-linear-equations-using-matrices.html
or determinants http://www.purplemath.com/modules/determs.htm
No, then
a. x-4y-6z=-34
b. -7x+5y-7z=-24
c. 5x-7y-z=-18
Multiply equation a by 7
a 7x-28y -42z=- 238 add b)
b.-7x+5y -7z=-24
d. -23y-49z=-362
take equation a, and c
a. x-4y-6z=-34
c. 5x-7y-z=-19
multiply a) by 5,
a. 5x-20y-30z=-170
c. 5x-7y-z=-19, subtract equation c from a,
e. -14y-29z=-151 and from d)
d) -23y-49z=-362
now we have two equations, two unknowns. geting messy.
multiply 3 by 23, and d) by 14, then subtract e from d, and you have one equation, one unknown.
(-29*23+14*49)z=-151*23+362*14
then z= ... and you go back up. By far, one of the three methods I mentioned first is much easier. Look into those links if your teacher has not broached the subjects.
or
http://hotmath.com/hotmath_help/topics/solving-systems-of-linear-equations-using-matrices.html
or determinants http://www.purplemath.com/modules/determs.htm
No, then
a. x-4y-6z=-34
b. -7x+5y-7z=-24
c. 5x-7y-z=-18
Multiply equation a by 7
a 7x-28y -42z=- 238 add b)
b.-7x+5y -7z=-24
d. -23y-49z=-362
take equation a, and c
a. x-4y-6z=-34
c. 5x-7y-z=-19
multiply a) by 5,
a. 5x-20y-30z=-170
c. 5x-7y-z=-19, subtract equation c from a,
e. -14y-29z=-151 and from d)
d) -23y-49z=-362
now we have two equations, two unknowns. geting messy.
multiply 3 by 23, and d) by 14, then subtract e from d, and you have one equation, one unknown.
(-29*23+14*49)z=-151*23+362*14
then z= ... and you go back up. By far, one of the three methods I mentioned first is much easier. Look into those links if your teacher has not broached the subjects.
boring !
x-4y-6z=-34 times 5
-7x+5y-7z=-24 times 4
+5 x -20 y - 30 z = -170
-28x +20 y - 28 z = -96
--------------------------add
-23 x + 0 -58 z = - 266
now the last 2
-7x+5y-7z=-24 times 7
+5x-7y-1z=-18 times 5
-49 x + 35 y - 49 z = -168
+25 x - 35 y - 5 z = -90
--------------------------add
-24 x + 0 -54 z = - 258
so
-23 x -58 z = - 266 times 24
-24 x -54 z = - 258 times 23
-552 x - 1392 z = -6384
-552 x - 1242 z = -5934
------------------------subtract
-150 z = -450
z = 3 whew now all yours
x-4y-6z=-34 times 5
-7x+5y-7z=-24 times 4
+5 x -20 y - 30 z = -170
-28x +20 y - 28 z = -96
--------------------------add
-23 x + 0 -58 z = - 266
now the last 2
-7x+5y-7z=-24 times 7
+5x-7y-1z=-18 times 5
-49 x + 35 y - 49 z = -168
+25 x - 35 y - 5 z = -90
--------------------------add
-24 x + 0 -54 z = - 258
so
-23 x -58 z = - 266 times 24
-24 x -54 z = - 258 times 23
-552 x - 1392 z = -6384
-552 x - 1242 z = -5934
------------------------subtract
-150 z = -450
z = 3 whew now all yours
5 answers