Does the system have no solution?
1 answer:
Answer:
Step-by-step explanation:
Notice that in the second equation, the coefficient of is
times greater than in the first. Therefore, by making the coefficient of
in the second equation
times greater than in the first, it's impossible to only isolate one variable.
Therefore, when , there are no solutions to this system.
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Answer:
Its definitley square to be precise so that's my answer, but its also a rhombus
Step-by-step explanation:
assuming a is 1, regardless, graph would always be square, which is a rhombus. hope square is correct :)
Answer:
Option A is correct (17,11).
Step-by-step explanation:
6x - 9y = 3
3x - 4y =7
it can be represented in matrix form as
A=
X=
B=
i.e, AX=B
or X= A⁻¹ B
A⁻¹ = 1/|A| * Adj A
determinant of A = |A|= (6*-4) - (-9*3)
= (-24)-(-27)
= (-24) + 27 = 3
so, |A| = 3
Adj A=
A⁻¹ = /3
A⁻¹ =
X= A⁻¹ B
X=
X=
X=
X=
x= 17, y= 11
solution set= (17,11).