Question

X - y = 2 4x – 3y = 11

Answer 1

Answer:

For x - y = 2

You need to find x or y in one of the equations and then substitute that into the other.

So we have;

x-y=2

4x-3y=11

We will take the first equation and find x;

x-y=2

add y to both sides;

x-y+y=2+y

x=2+y

Now we take that answer and substitute it forx in the other equation;

 

4(2+y)-3y=11

8+4y-3y=11

8+y=11

y=3

Now we have what y equals, so we use it in the first equation to find x;

x-3=2

x=5

So we have;

x=5; y=3

Hope you understand!

=)

And for 4x – 3y = 11

Multiply the first equation by 2 and the second by 3 so that there are the same number of y's in each:

8x - 6y = 22    ...(3)

30x + 6y = -3  ...(4)

 Now add (3) and (4) term by term:

38x + 0 = 19

or

38x = 19

or x = 1/2

Put this back into equation (1)

4*(1/2) - 3y = 11

or

2 - 3y = 11

Subtract 2 from both sides:

-3y = 9

 Divide both sides by -3

y = -3

Answer 2

Answer:

The answer is one solution, the lines intersect at point (5,3)

Step-by-step explanation:

The method I used to solve this is substitution.

1.  First I solved for x in the equation 4x – 3y = 11. Which is x = $\frac{3y+11}{4}$

2.  Second solved for y in the equation x - y = 2. Which is y = x-2

3. Then substitute the y in the equation x = $\frac{3y+11}{4}$  for y = x-2 to find the value of x. Which is x = 5

4. Lastly plug in 5 for x in the equation y = x-2. Which is three

The answer is one solution, the lines intersect at point (5,3)

Explanation for each step.

1. 4x - 3y = 11

        +3y (add 3y to both sides of equation.)

   4x = (3y + 11 )/4 divide both sides by 4, to get x alone

  x = $\frac{3y+11}{4}$

2. x - y = 2.

        + y (add y to both sides of equation.

    x = y +2

3. x = $\frac{3y+11}{4}$

       = $\frac{3(x-2)+11}{4}$   (substitute)

       = $\frac{3x -6 + 11}{4}$    (add -6 and 11)

4(x) = ($\frac{3x + 5}{4}$)4 multiply both side by 4

  4x = 3x + 5

        -3x subtract -3x

      x = 5

4. y = x-2

      = 5 - 2 substitute

    y = 3

Side note: this took waaaay too long to complete, yet i hope this helped :)