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mamaluj [8]
3 years ago
15

What is the solution of the following linear system of equations? 4y=−5x−2, x−4y=14

Mathematics
1 answer:
Lisa [10]3 years ago
7 0

Answer:

X = 2 and Y = -3

Step-by-step explanation:

4y = -5x - 2

x - 4y = 14

5x + 4y = -2....... equation 1

x - 4y = 14........... equation 2

<u>Using Elimination </u><u>Method</u>

Add Equation 1 to Equation 2

6x = 12

Divide both side by 6

X= 2

Put X= 2 into Equation 1

5(2) + 4y = -2

10 + 4y = -2

4y = -2 - 10

4y = -12

Divide both side by 4

Y = -3

Therefore

x = 2

y = -3

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Cotx = cosx/sinx
cscx = 1/sinx

If we plug these into your equation we get:
(cosx/sinx)/(1/sinx) - 1

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8 0
3 years ago
In an arithmetic sequence, t4 + t5 + t6 = 300, and t15 +t16+ t17 = 201. Find t18.
stiks02 [169]
In an arithmetic sequence:
Tn=t₁+(n-1)d

t₄=t₁+(4-1)d=t₁+3d
t₅=t₁+(5-1)d=t₁+4d
t₆=t₁+(6-1)d=t₁+5d

t₄+t₅+t₆=(t₁+3d) +(t₁+4d)+(t₁+5d)=3t₁+12d

Therefore:
3t₁+12d=300    (1)


t₁₅=t₁+(15-1)d=t₁+14d
t₁₆=t₁+(16-1)d=t₁+15d
t₁₇=t₁+(17-1)d=t₁+16d

t₁₅+t₁₆+t₁₇=(t₁+14d)+(t₁+15d)+(t₁+16d)=3t₁+45d

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3t₁+45d=201    (2)

With the equations (1) and (2) we make an system of equations:

3t₁+12d=300
3t₁+45d=201

we can solve this system of equations by reduction method.

  3t₁+12d=300
-(3t₁+45d=201)
-----------------------------
       -33d=99      ⇒d=99/-33=-3

3t₁+12d=300
3t₁+12(-3)=300
3t₁-36=300
3t₁=300+36
3t₁=336
t₁=336/3
t₁=112

Threfore:

Tn=112+(n-1)(-3)
Tn=112-3n+3
Tn=115-3n

Now, we calculate T₁₈:

T₁₈=115-3(18)=115-54=61

Answer: T₁₈=61




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