SOLUTION: Please solve the system of equation for all variables using a matrix.
{{{-2x+y=-4}}}
{{{x+3y=9}}}
My solution
2 1=-4
1 3=9
r2 reverse r1
1 3=9
2 1=-4
r2* -2
1 3=9
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-> SOLUTION: Please solve the system of equation for all variables using a matrix.
{{{-2x+y=-4}}}
{{{x+3y=9}}}
My solution
2 1=-4
1 3=9
r2 reverse r1
1 3=9
2 1=-4
r2* -2
1 3=9
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Question 146084This question is from textbook Finite Mathematics
: Please solve the system of equation for all variables using a matrix.
My solution
2 1=-4
1 3=9
r2 reverse r1
1 3=9
2 1=-4
r2* -2
1 3=9
0 -1=-6
r2* 2
1 3=9
0 1=-12
r2*-3+r1
1 0=-6
0 1=-12
x= -6 and y= -12
Please help I'm trying to get these steps, but i think my mutiplcation is off somewhere. Thanks in advance. This question is from textbook Finite Mathematics
You can put this solution on YOUR website! Please solve the system of equation for all variables using a matrix.
-2x+y = -4
x +3y = 9
------------------
My solution
2 1=-4
1 3=9
-----------
r2 reverse r1
1 3=9
2 1=-4
------------
r2* -2
1 3=9
0 -1=-6
Comment: You cannot subtract 2 from the x-term, 2 from the y-,and 2 from
the constant and still maintain the equation.
What should you do?
You should subtract 2*(1st row) from the 2nd row to get
0 -5 = -22
Then continue on.
================
Cheers,
Stan H.
--------------
r2* 2
1 3=9
0 1=-12
r2*-3+r1
1 0=-6
0 1=-12
x= -6 and y= -12
You can put this solution on YOUR website!
Hi, you are literally trying to get the identity on the left hand side of the matrix.
You dont have to swap the rows but if you do then you will have:
1 3 = 9
-2 1=-4
then to get a zero in row 2 you can add 2 row1:
1 3=9
0 7=14
then we need a 1 where the 7 is and therefore we divide row 2 by 7:
1 3=9
0 1=2
Row 1 minus 3 row 2:
1 0=3
0 1=2
Therfore you read off your solution that x=3 and y=2 because column 1 relates to x and column 2 is y.
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