Solve the following simultaneous linear equations: (x-2)/5 = (1-y)/4 and 26x+3y+4=0  

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve the set of equations:

(x - 2)/5 = (1 - y)/4 ...(1)

26x + 3y + 4 = 0 ...(2)

(x - 2)/5 = (1 - y)/4

=> 4x - 8 = 5 - 5y

=> 4x + 5y - 13 = 0 ...(3)

From (3)

x = (13 - 5y)/4

Substitute in (2)

26(13 - 5y)/4 + 3y + 4 = 0

338 - 130y + 12y + 16 = 0

=> 118y = 354

=> y = 3

x = (13 - 5*3)/4 = -2/4 = -1/2

The solution of the given system of equations is x = -1/2 and y = 3

misshakunamatata's profile pic

misshakunamatata | Student, College Freshman | (Level 1) eNoter

Posted on

I agree with justaguide  and hery. When it says linear,it means the two equations are equal. And here is my step by step solution

Given: Eq.1(x-2)/5=(1-y)/4.........Eq.2 26x+3y=4

When you first use the Eq. 1.this is how it looks like

Given (x-2)/5=(1-y)/4

x-2=[(1-y)/4]5

x-2=(5-5y)/4

x=[(5-5y)/4]+2

x=[(5-5y)/4]+8/4

x=(13-5y)/4

Substitute the value of x to the Eq 2

Always Equate the given Equation to zero

Given 26x+3y+4=0

26[(13-5y)/4]+3y+4=0

(338-130y)/4 +3Y+4=0

(338-130y+12y+16)/4=0

(354-118y)/4=0

354/4=118y/4

354=118y

354/118=118y/118

y=3

Substitute the value of y to any equation to get the value of  x

Using Eq. 1

(x-2)/5=(1-y)/4

(x-2)/5=(1-3)/4

(x-2)/5=-2/4

x-2=(-2/4)5

x-2=-10/4

x=-5/2 + 2

x=-5/2 +4/2

x=-1/2

USing Eq 2

26x+3y+4=0

26x+3(3)+4=0

26x+9+4=0

26x+13=0

26x=-13

26x/26=-13/26

x=-13/26

x=-1/2 (always simplify to lowest term)

 

therefore x=-1/2 and y=3

 

I hope it will help you.Step by step procedure is a great help especially when you are just starting.

 

 

 

 

hery's profile pic

hery | College Teacher | (Level 1) Honors

Posted on

I agree with justaguide, the answer is x = -1/2 and y = 3

The step for solving the problem in my way is:

We have two equations:

(1) 1/5(x-2) = 1/4 (1-y)

(2) 26x + 3y + 4 = 0

Step #1: Process the (1) equation:

1/5 (x-2) = 1/4 (1-y)

=> (x-2)/5 = (1-y)/4

=> 4(x-2) = 5(1-y)

=> 4x - 8 = 5 - 5y

=> 4x + 5y -8 - 5 = 0

=> 4x + 5y - 13 = 0

 

Gime Times 3 to (1) => 3 (4x+5y-13) =0

=> 12x + 15y - 39 = 0 ................(1*)

Give Times 5 to (2) => 5 (26x +3y + 4) = 0

=> 130x + 15y + 20 = 0  ......(2*)

 

(1*) and (2*)

12x + 15y - 39 = 0

130x + 15y + 20 = 0  -

__________________

-118 x - 59 = 0  => - 118 x = 59

=> x = 59/-118 = -1/2

Substitude x = -1/2 to either (1) or (2)

Here I choose to substitute x = -1/2 to (1)

=> 4x + 5y - 13 = 0

=> 4(-1/2) + 5y - 13 = 0

=> -2 + 5y - 13 = 0

=>  5y - 15 = 0

=>  5y = 15

=>  y = 15/5 = 3

Thus, the answer is x = -1/2 & y = 3

 

 

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