How do you solve #4x-12y=5# and #-x+3y=-1# using substitution?

Question

Nathan Axel · Accepted Answer

No soluion.

In substitution method we get the value of one variable in term of other from one equation and then put it in another equation. Here in second equation, transposing #x# term to RHS and constant term to LHS gives us #3y+1=x# or #x=3y+1#. Now putting this in first equation we get #4(3y+1)-12y=5# or #12y+4-12y=5# or #4=5#, which is not true and hence we have no solution. Additional fact - This means that if we draw two lines on a Cartesian plane, they lead too two parallel lines and hence no solution. Had it been #0=0#, it would have been two coincident lines and hence infinite solutions.

Genesis Allen · Answer

undefined To solve the system of equations $4x - 12y = 5$ and $-x + 3y = -1$ using substitution:

1. Solve one of the equations for one variable in terms of the other variable.
2. Substitute the expression found in step 1 into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found in step 3 back into one of the original equations to find the value of the other variable.
5. Write the solution as an ordered pair (x, y).

Let's solve the system using these steps:

From the second equation, solve for $x$:
$-x + 3y = -1$  
$x = 3y + 1$

Substitute $x = 3y + 1$ into the first equation:
$4(3y + 1) - 12y = 5$  
$12y + 4 - 12y = 5$  
$4 = 5$

This equation is contradictory, indicating that there is no solution to the system of equations.