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Mrrafil [7]
3 years ago
7

Solve for 2 unknowns : 3a+4b=9 ; 2a+2b=6

Mathematics
1 answer:
BARSIC [14]3 years ago
3 0
Let us take the second equation first
2a + 2b = 6
Dividing both sides by 2 we get
a + b = 3
a = 3 - b
Putting the value of a in the first equation we get
3a + 4b = 9
3(3 - b) + 4b = 9
9 - 3b + 4b = 9
b = 9 - 9
   = 0
Now putting the value of b in the second equation we get
a + b = 3
a + 0 = 3
a = 3
So the value of the unknown variable a is 3 and the value of the unknown variable b is 0.

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Answer:

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Step-by-step explanation:

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-> nxm multiplied by pxq gives nxq eg 3x2 mul 2x2 gives 3x2

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So considering the above facts :-

a) AB+C = [(2x2)(2X2)]+(2x2)=(2x2)+(2x2)=2x2

b) 3GF =3[(3x3)(3x3)]=3(3x3)=3x3

c)CK+B= [(2x2)(2x3)]+(2x2)=(2x3)+(2x2) which is not possible (see point 1)

d)CK+H =as CK is (2x3) so (2x3)+(2x3)=(2x3)

e)EMC=[(3x3)(3x2)](2x2)=(3x2)(2x2)=3x3

f)GLH=[(3x3)(3x2)](2x3)=(3x2)(2x3)=3x3

g)HLG=[(2x3)(3x2)](3x3)=(2x2)(3x3)= not possible(see point 2)

h)2EL+5MB=2[(3x3)(3x2)]+5[(3x2)(2x2)]=(3x2)+(3x2)=3x2

   

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Answer:

\begin{bmatrix}\mathrm{Solution:}\:&\:x\le \frac{1200}{499}\:\\ \:\mathrm{Decimal:}&\:x\le \:2.40480\dots \\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:\frac{1200}{499}]\end{bmatrix}

Step-by-step explanation:

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