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lara31 [8.8K]
3 years ago
14

-x-3y=14 x=4y solve using substitution

Mathematics
2 answers:
kherson [118]3 years ago
6 0
Solve both equations for one variable.  Since the second equation is already solved for x, we'll solve the other one for x.

x=-14-3y and x=4y are our two equations.  

Now we can assume that if x = -14-3y and x=4y, we can assume that -14-3y = 4y.  If we solve for y, we get that y=-2.  Plug in y=-2 into either one of the equations, I'll use x=4y.  x=4(-2) so [/tex]x=-8[/tex].  Know we know what x and y are, so we can put it in an orderned pair:  (-8,-2)
vlabodo [156]3 years ago
5 0
Using
 f1(x)= -x=3t =14

with f2(x) as x=4y

f1(f2)= -x-3y=14
F1(4y)= -(4y)-3y=14
           = -4y-3y=14
           = - 7y=14
           = y= -2

Using the value of y, solve fo rx by entering the value in the original form of f1 or f2

<span>-x-3y=14</span>
-x-3(-2)=14
-x+6=14
x=-8

OR
x=4y
  =4(-2)
  = -8

Thus the solution is (x,y) = (-8,-2)
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Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

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plug into 4x+4y=32 to solve for y

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Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

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