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worty [1.4K]
4 years ago
11

match each system to the number the first equation can be multiplied by to eliminate the x-terms when adding the second equation

Mathematics
2 answers:
Ratling [72]4 years ago
5 0

Answer:

1. 10x-4y=-8 matches: 1/2

-5x+6=10

2. -2x+6y=3 matches: 2

4x+3y=9

3. 3x-8y=1 matches: -2

6x+5y=12

4. -8x+10y=16 matches: -1/2

-4x-5y=13

seraphim [82]4 years ago
3 0

Part a: -2 is used to eliminate the x-terms when adding with the second equation.

Part b: -\frac{1}{2} is used to eliminate the x-terms when adding with the second equation.

Part c: \frac{1}{2} is used to eliminate the x-terms when adding with the second equation.

Part d: 2 is used to eliminate the x-terms when adding with the second equation.

Explanation:

Part a: The equations are 3 x-8 y=1 and 6 x+5 y=12

To eliminate the x-terms from both the equations, let us multiply -2 with the first equation and hence it becomes -6x+16y=-2

Adding the two equation, we get,

-6x+16y+6 x+5 y=-2+12

Simplifying, we get,

21y=10

Thus, the x-terms are eliminated when adding the equations.

Hence,-2 is used to eliminate the x-terms when adding with the second equation.

Part b: The equations are -8 x+10 y=16 and -4 x-5 y=13

To eliminate the x-terms from both the equations, let us multiply -\frac{1}{2} with the first equation and hence it becomes 4 x-5 y=-8

Adding the two equation, we get,

4x-5y-4x-5y=-8+13

Simplifying, we get,

0=5

Thus, the x-terms are eliminated when adding the equations.

Hence, -\frac{1}{2} is used to eliminate the x-terms when adding with the second equation.

Part c: The equations are 10 x-4 y=-8 and -5 x+6 y=10

To eliminate the x-terms from both the equations, let us multiply \frac{1}{2} with the first equation and hence it becomes 5x-2y=-4

Adding the two equation, we get,

5x-2y-5x+6y=-4+10

Simplifying, we get,

4y=6

Thus, the x-terms are eliminated when adding the equations.

Hence, \frac{1}{2} is used to eliminate the x-terms when adding with the second equation.

Part d: The equations are -2 x+6 y=3 and 4 x+3 y=9

To eliminate the x-terms from both the equations, let us multiply 2 with the first equation and hence it becomes -4x+12y=6

Adding the two equation, we get,

-4x+12y+4x+3y=6+9

Simplifying, we get,

15y=15

Thus, the x-terms are eliminated when adding the equations.

Hence, 2 is used to eliminate the x-terms when adding with the second equation.

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Normally, I would pass right by this question, because it is so complicated
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to tread.

The answer is a number, that represents "after how many hours".  So we
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clear on when the beginning of time is, so I need to decide.  I'm going to
say that the time when the bus leaves is the beginning of time.  And I'm
going to call  ' H '  the number of hours AFTER the bus leaves.

-- The bus covers 50 miles every hour.  So 'H' hours after the bus leaves,
the bus is  50H  miles away from the city, in some direction.

-- The car leaves 2 hours later.  So 'H' hours after the bus leaves, the car
has only been traveling for  (H-2)  hours.

-- The car covers 70 miles every hour.  So 'H' hours after the bus leaves,
the car is  (70)x(H-2)  miles away from the city, in the OTHER direction.

-- Since the bus and the car are moving in opposite directions, the distance
between the bus and the car is the SUM of the distances each one covers.

     'H' hours after the bus leaves, they are  (50H) plus (70)(H-2) miles apart.

What we need to figure out is:  When is that distance 800 miles ?


The distance between them at 'H' hours after the bus leaves is

                                                            (50H) plus (70)(H-2) miles

Eliminate the first parentheses:             50H  +  (70)(H-2)

Eliminate the rest of the parentheses:   50H  +  70H - 140

Combine the terms with 'H' in them:         120H - 140

So here's the equation:                            120H - 140 = 800

Add  140  to each side:                            120H          =  940

Divide each side by  120 :                             H  =  7.8333 hours
                                                                           =  7 and 5/6 hours
                                                                           =  7 hours and 50 minutes
                                                                                 after the bus leaves.

Check:

Assume that  H = 7 hrs 50 minutes  (7 and 5/6 hours)
The bus has covered  (50 x 7-5/6) = 391 and 2/3 miles.
 
The car has been traveling for only  5 and 5/6 hours
The car has covered   (70) x (5-5/6) =  408 and 1/3 miles

How far apart are they ?   (391-2/3) + (408-1/3) =  800 miles !           YAY!   


Note:  If the book or the homework sheet wants to count the hours
after the CAR leaves, then the answer is  5-5/6 hours, not 7-5/6 . 

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