i) Solve the simultaneous equations 4x + 3y + 4 = 0, 3x - 12y + 22 = 0.  Could you please teach me step by step, so that I can understand easily and without any problem? Thank you. :)

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pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

4x + 3y + 4 = 0, 3x - 12y + 22 = 0.

I think the easiest way to do this is by addition.  So we need to make it so two of the variables (one each from each equation) will come to 0 when we add them.

I think the best way to do this is to multiply the first equation by 4 because then we will have 12y to cancel the -12y in the other equation.

So we multiply by 4 and we have

16x + 12y + 16

Now we add that equation to the other.  We add the like terms together.

16x + 3x is 19x

12y + (-) 12 y  is 0

16 + 22 is 38.

So we have 19x + 38 + 0

Move the 38 to the other side, we have

19x = -38

x = -2

Substitute that back into the first equation to solve for y.

4(-2) + 3y +4 = 0

-8 +3y +4 + 0

3y = 8 - 4

3y = 4

y = 4/3

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The given simultaneous equations are:

4x + 3y + 4 = 0   ...    (1)

3x - 12y + 22 = 0   ... (2)

Multiplying equation (1) by 4 we get:

16x + 12y + 16 = 0   ...    (3)

adding equation (2) and (3) we get:

3x + 16x - 12y + 12y + 22 + 16 = 0

19x + 38 = 0

19x = -38

Therefore:

x = -38/19 = -2

Substituting this value of x in equation (1) we get:

4*(-2) + 3y + 4 = 0

-8 + 3y + 4 =0

3y = 4

Therefore:

y = 4/3

Answer:

x = -2, and y = 4/3

Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

(1)  4x + 3y + 4 = 0
(2)  3x - 12y + 22 = 0

Multiply (1) by 4

(3)  16x + 12y + 16 = 0

Add (3) and (2)

19x + 38 = 0

Now its single variable and you can solve for x. Plug this value into one of the equations (1) or (2) and solve for y.

princessenotes's profile pic

princessenotes | Student, Undergraduate | (Level 1) Honors

Posted on

16x + 12y + 16

16x + 3x is 19x

12y + (-) 12 y  is 0

16 + 22 is 38.

19x + 38 + 0

19x = -38

x = -2

 

To solve for y:

4(-2) + 3y +4 = 0

-8 +3y +4 + 0

3y = 8 - 4

3y = 4

y = 4/3

 

I hope I helped you. Bye!

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

4x+3y+4=0..............(1)

3x-12y+22=0...........(2) are the two equations given to be solved for x and y.

From eq(1) we get: 3y = -(4x+4)  Or

y = -(4x+4)/3. Substituting in eq(2) y = -(4x+4)/3, we get:

3x-12[-(4x+4)/3]+22=0. Or

3x+4(4x+4)+22 = 0. Or

19x +38 = 0 Or

x = -38/19 = -2. Substituting this value of x = -2 in (1),

4(-2)+3y+4= 0. Or

3y = 4 or y = 4/3

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