Answer:
Step-by-step explanation:
Given the simultaneous equation 2p - 3q = 4 and 3p + 2q = 9, to get the value of p and q we will use elimination method.
2p - 3q = 4 ...................... 1 * 3
3p + 2q = 9 ..................... 2 * 2
Multiplying equation 1 by 3 and 3 by 2:
6p - 9q = 12
6p + 4q = 18
Subtracting both equation
-9q-4q = 12-18
-13q = -6
q = -6/-13
q = 6/13
Substituting q = 6/13 into equation 2
2p - 3(6/13) = 4
2p - 18/13 = 4
2p = 4+18/13
2p = (52+18)/13
2p = 70/13
p = 70/26
p = 35/13
<em>Hence p = 35/13 and q = 6/13</em>
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<em>b) </em>If if 223ₓ = 87 find x
Using the number base system and converting 223ₓ to base 2 will give us;
223ₓ = 2*x² + 2*x¹ + 3*x⁰
223ₓ = 2x²+2x+3
Substituting back into the equation, 2x²+2x+3 = 87
2x²+2x+3-87 = 0
2x²+2x-84 = 0
x²+x-42 = 0
On factorizing:
(x²+6x)-(7x-42) = 0
x(x+6)-7(x+6) = 0
(x+6)(x-7) = 0
x+6 = 0 and x-7 = 0
x = -6 and 7
<em>Hence the value of x is 7 (neglecting the negative value)</em>
Y=-13/2x+43/2 I hope this helped :3
Answer:
y=2x+4
Step-by-step explanation:
the slop is 2 hense 2x, and the y intercept is 4
Answer:
3 games
Step-by-step explanation:
The computation of the maximum number of games that he could buy is shown below:
= Amount available - already spent
= $75 - $28
= $47
If each game cost $15
So, the maximum no of games is
= $47 ÷ $15
= 3.133
= 3 games
A(b-c)=ab-ac
(1/7)(1/3h-1/2)=(1/7)(1/3h)-(1/7)(1/2)=1/21h-1/14
now we gots
6 and 1/7+1/21h-1/14
rearange
6 and 1/7-1/14+1/21h
make common denom which is 14
6 and 2/14-1/14+1/21h
6 and 1/14+1/21h