4. Substituting for y:
.. 7x -(6 -2x) = 3
.. 9x = 9 . . . . . . . . . add 6, collect terms
.. x = 1
.. y = 6 -2*1 = 4
(x, y) = (1, 4)
5. Substituting for x:
.. 3(y -4) +y = 12
.. 4y = 24 . . . . . . . add 12, collect terms
.. y = 6
.. x = 6 -4 = 2
(x, y) = (2, 4)
6. Substituting for x:
.. 4(y -5) -y = 4
.. 3y = 24 . . . . . . . add 20, collect terms
.. y = 8
.. x = 8 -5 = 3
(x, y) = (3, 8)
7. y = 2x -6 . . . . . add 2x to the first equation
Substituting for y:
.. 5x -(2x -6) = 9
.. 3x = 3 . . . . . . . add -6, collect terms
.. x = 1
.. y = 2*1 -6 = -4
(x, y) = (1, -4)
8. x = 2 -3y . . . . . add -3y to the first equation
Substituting for x:
.. y -(2 -3y) = -6
.. 4y = -4 . . . . . . . add 2, collect terms
.. y = -1
.. x = 2 -3*(-1) = 5
(x, y) = (5, -1)
Answer:
hello :
f(x) = -x+0.25
g(x) = x²+3x+2
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f(-3.5) = 3.75
g(-3.5) = 3.75
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f(-0.5) =0.75
g(-0.5) = 0.75
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conclusion : answer : A) x = −3.5 and x = −0.5
Step-by-step explanation:
Hope this helps:)
Answer:
Yes
With an extra 3000CHF
Step-by-step explanation:
In year 2005, the exchange rate was 1$ = 1.15CHF
The student came in with $20,000.
The amount he had in CHF(swiss franc) = ?
$1 = 1.15C
$20,000 = x CHF
X = (20000 × 1.15) / 1
X = 23,000CHF
When he came to Zürich in 2005, he has 23,000 swiss franc.
But the exchange rates changed in 2006 to C1.30 against the dollar.
$1 = 1.30CHF
$20,000 = xCHF
X = 26,000CHF
In 2006, he would've had 26,000CHF.
The difference between what he would've had in 2006 against 2005 is
26,000CHF - 23,000CHF = 3000CHF
He would've had an extra 3000CHF if he came in the year 2006.
Answer:
m + 8
Step-by-step explanation:
you start with the original number of marbles, m, and you ADD 8 more to the group.
Original number (m) + 8 more
M = 8