Answer:
(i)unique solution
Explanation:
We solve for x1 thus in the first equation:
x1-3x3=-3
x1-9=-3
x1=-3+9
x1= 6
We solve for a thus in the second equation:
2x1+ax2-x3=-2
2+a2-x3=-2
a2-x3=-4
a2=-4+x3
Answer:
(-7, -12)
Step-by-step explanation:
4x-3y=8
5x-2y=-11
Is there any of the like terms can be added and the result will be 0? No, so we have to multiple one OR both of the equations to make that one number do that.
(I will try to remove the y like terms so i will multiple both of them by the opposite so both of the ys will be 6)
2(4x-3y=8)
-3(5x-2y=-11)
8x-6y=16
-15x+6y=33
(now the easy part… cancel the 6s and add the equations)
8x+(-15x)=-7x
16+33=49
-7x=49
(divide 49 by -7)
x=-7
Replace x in any of the equations and you’ll get the y value.
4x-3y=8
4(-7)-3y=8
-28-3y=8
-3y=36
y=12
Threfore, there is one solution which is….. (-7,-12)
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