Question 672662
There are many ways to do this, but here is one of them.


Convert 


4a+5b-6c=2
-3a-2b+7c=-15
-a+4b+2c=-13


to matrix form to get



<pre>
4	5	-6	2
-3	-2	7	-15
-1	4	2	-13
</pre>


Then use row reduction


<pre>
4	5	-6	2
-3	-2	7	-15
-1	4	2	-13


1	1.25	-1.5	0.5     	0.25*R1
-3	-2	7	-15	
-1	4	2	-13	


1	1.25	-1.5	0.5	
0	1.75	2.5	-13.5   	R2 + (3)*R1
-1	4	2	-13	


1	1.25	-1.5	0.5	
0	1.75	2.5	-13.5	
0	5.25	0.5	-12.5   	R3 + (1)*R1


1	1.25	-1.5	0.5	
0	1	1.4286	-7.7143 	0.571429*R2
0	5.25	0.5	-12.5	


1	0	-3.2857	10.1429 	R1 + (-1.25)*R2
0	1	1.4286	-7.7143	
0	5.25	0.5	-12.5	


1	0	-3.2857	10.1429	
0	1	1.4286	-7.7143	
0	0	-7	28      	R3 + (-5.25)*R2


1	0	-3.2857	10.1429	
0	1	1.4286	-7.7143	
0	0	1	-4      	-0.142857*R3


1	0	0	-3      	R1 + (3.28571)*R3
0	1	1.4286	-7.7143	
0	0	1	-4	


1	0	0	-3	
0	1	0	-2      	R2 + (-1.42857)*R3
0	0	1	-4	


</pre>


The right hand column is -3, -2, -4


These 3 values correspond to a, b, and c respectively.


So the solutions are...


a = -3


b = -2


c = -4


They form the ordered triple (-3, -2, -4)