Solving Linear System of Equations: Elimination

In summary, the equations have the same coefficients but y is there because it's trying to make the coefficients of x the same.
  • #1
ilovecake
2
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Homework Statement



A general system of linear equations is
ax+by=e
cx+dy=f

where a,b,c,d,e,f are constant values.

a)Use elimination to solve for x and y in terms of a,b,c,d,e,f.

b)Are there any values that a,b,c,d,e,f cannot have?

The Attempt at a Solution



For part A, I got x=(e-f)/(a-c), assuming using subtraction for elimination, but in the answers it says x=(de-bf)/(ad-bc). Why are the coefficients of y there?

For part B, I think ad=/=bc because then both equations would cancel out, and it is either the exact same equation or has the same slope but parallel and never intersecting.

for example, 15x+9y & 5x+3y
 
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  • #2
ilovecake said:

Homework Statement



A general system of linear equations is
ax+by=e
cx+dy=f

where a,b,c,d,e,f are constant values.

a)Use elimination to solve for x and y in terms of a,b,c,d,e,f.

b)Are there any values that a,b,c,d,e,f cannot have?


The Attempt at a Solution



For part A I got x=(e-f)/(a-c), assuming using subtraction for elimination, but in the answers it says x=(de-bf)/(ad-bc). Why is the coefficient of y there?

For part B I believe ad=/=bc because if ad=bc, then the x and y of both equations would have the same coefficients meaning it's the exact same equation or same slope but parallel and never intersecting.

You need to show your work---how you got the answer, not just the answer itself.

RGV
 
  • #3
using elimination by subtraction, solving for x:
ax+by=e
cx+dy=f

(ax+by)-(cx+dy)=(e-f)
ax-cx +by-dy=(e-f)
(a-c)x=(e-f)
x=(e-f)/(a-c)

Nevermind, I just realized that it's to make the coefficients of y the same.

(ad)x+[STRIKE](bd)y[/STRIKE]=(de)
(cb)x+[STRIKE](bd)y[/STRIKE]=(bf)
 
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FAQ: Solving Linear System of Equations: Elimination

What is the purpose of solving a linear system of equations using elimination?

Solving a linear system of equations using elimination allows us to find the values of the variables that satisfy all of the equations in the system. This is useful in many real-life situations, such as determining the intersection point of two lines or finding the solution to a system of equations representing a financial problem.

How does the elimination method work?

The elimination method involves eliminating one variable from the system by adding or subtracting the equations in the system. This is done by manipulating the equations to create a new equation with one less variable, which can then be solved to find the value of the eliminated variable. This process is repeated until all variables have been eliminated, resulting in a set of values that satisfy all equations in the system.

Can all linear systems of equations be solved using elimination?

Yes, all linear systems of equations can be solved using elimination. This method is particularly useful for systems with two equations and two variables, but it can also be used for systems with more equations and variables. However, as the number of variables increases, the elimination process can become more complex and time-consuming.

Are there any limitations to using elimination to solve linear systems of equations?

One limitation of the elimination method is that it can only be used for systems of linear equations. If the equations in the system are not linear, then other methods, such as substitution or graphing, must be used. Additionally, if the system has infinitely many solutions or no solutions, the elimination method may not be able to find a solution.

Is there a specific order in which equations should be eliminated?

No, there is no specific order in which equations should be eliminated. However, it is often helpful to choose the variable that is easiest to eliminate first, such as one with a coefficient of 1 or -1. This can make the elimination process more efficient and less prone to errors.

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