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Rainbow [258]
3 years ago
13

Solve by substitution 3x+y=3 and 7x+2y=1

Mathematics
1 answer:
balu736 [363]3 years ago
5 0

3x + y = 3

7x + 2y = 1

First isolate one of the variables (x or y) in one of the equations.

Isolate "y" in the first equation(because it is the easiest to isolate) and substitute it into the second equation.

3x + y = 3      Subtract 3x on both sides

3x - 3x + y = 3 - 3x

y = 3 - 3x


7x + 2y = 1

7x + 2(3 - 3x) = 1     [since y = 3 - 3x, you can substitute (3-3x) for "y"]

Multiply/distribute 2 into (3 - 3x)

7x + (3(2) - 3x(2)) = 1

7x + 6 - 6x = 1

x + 6 = 1     Subtract 6 on both sides

x = -5


Now that you know "x", substitute it into one of the equations (I will do both)

3x + y = 3

3(-5) + y = 3       [since x = -5, you can plug in -5 for "x"]

-15 + y = 3    Add 15 on both sides

y = 18


7x + 2y = 1

7(-5) + 2y = 1

-35 + 2y = 1     Add 35 on both sides

2y = 36     Divide 2 on both sides

y = 18


x = -5, y = 18  or  (-5, 18)

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A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f
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Answer:

2580 ft-lb

Step-by-step explanation:

Water leaks out of the bucket at a rate of \frac{0.15 \mathrm{lb} / \mathrm{s}}{1.5 \mathrm{ft} / \mathrm{s}}=0.1 \mathrm{lb} / \mathrm{ft}

Work done required to pull the bucket to the top of the well is given by integral

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=46-0.1x

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a=0 \text { and } b=60

Find the work done as,

W=\int_{a}^{b} F(x) d x

=\int_{0}^{60}(46-0.1 x) dx

&\left.=46x-0.05 x^{2}\right]_{0}^{60}

=(2760-180)-0[

=2580 \mathrm{ft}-\mathrm{lb}


Hence, the work done required to pull the bucket to the top of the well is 2580 \mathrm{ft}- \mathrm{lb}

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An exam worth 145 points contains 50 questions. Some of the questions are worth two points and some are worth five points. How m
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A*5+b*2=145     
a    +b   =50       | * -2

5a   +2b=145
-2a  -2b = -100
----------------------
5a-2a    = 145-100
3a=  45
a=45:3
a=15 questions are worth 5 p
b=50-15=35 questions worth 2 p

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