How to Solve bu using elimination method ?

Question

Hazel Anglin · Accepted Answer

#x=-11/2,y=-9/2# Substituting #a=1/(x+y),b=1/(x-y)# so we have the System #5a+2b=3# #20a-3b=1# Multiplying the first equation by #-4# and adding to the second we get #b=-1# so #a=-1/10# Then we have #1=-x+y# #-10=x+y# from here we get #x=-11/2# #y=-9/2#

Genesis Allen · Answer

#x = 3 and y = 2#

#5/(x + y) + 2/(x - y)=3#

#20/(x + y)-3/(x - y)= 1#

Let;

#1/(x + y) = a and 1/(x - y) = b#

Consequently,

#5/(x + y) + 2/(x - y)=3#

#5(1/(x + y)) + 2(1/(x - y))=3#

#5a + 2b = 3 - - - eqn1#

In the same way.

#20/(x + y)-3/(x - y)= 1#

#20(1/(x + y))-3(1/(x - y)) = 1#

#20a - 3b = 1 - - - eqn2#

Applying the Elimination Process!

#5a + 2b = 3 - - - eqn1#

#20a - 3b = 1 - - - eqn2#

Multiply #eqn1# by #3# and #eqn2# by #2#

#3(5a + 2b = 3)#

#2(20a - 3b = 1)#

#15a + 6b = 9 - - - eqn3#

#40a - 6b = 2- - - eqn4#

Adding both #eqn3 and eqn4# together..

#(15a + 40a) + (6b + (-6b))= 9 + 2#

#55a + 6b - 6b = 11#

#55a = 11#

#a = 11/55#

#a = 1/5#

Substituting the value of #a# into #eqn1#

#5a + 2b = 3 - - - eqn1#

#5(1/5) + 2b = 3#

#cancel5(1/cancel5) + 2b = 3#

#1 + 2b= 3#

#2b = 3 - 1#

#2b = 2#

#b = 2/2#

#b = 1#

But;

#1/(x + y) = a and 1/(x - y) = b#

#a = 1/(x + y)#

#1/5 = 1/(x + y)#

multiplication by cross;

#1(x + y) = 1(5)#

#x + y = 5 - - - eqn5#

Similarly..

#b = 1/(x - y)#

#1 = 1/(x- y)#

#1/1 = 1/(x- y)#

Cross multiplying;

#1(x - y) = 1(1)#

#x - y = 1 - - - eqn6#

Solving simultaenously again..

#x + y = 5 - - - eqn5#

#x - y = 1 - - - eqn6#

Using Elimination Method!

Adding #eqn5 and eqn6# together;

#(x + x) + (y +(-y)) = 5 + 1#

#2x + y - y = 6#

#2x = 6#

#x = 6/2#

#x = 3#

Substituting the value of #x# into #eqn6#

#x + y = 5 - - - eqn5#

#3 + y = 5#

Collecting like terms;

#y = 5 - 3#

#y = 2#

Hence;

#x = 3 and y = 2#

Hazel Anglin · Answer

#x=3,y=2#

#5/(x+y)+2/(x-y)=3# #20/(x+y)-3/(x-y)=1#

Let a be x+y and b be x-y.

#:.5/a+2/b=3#----------(1)

#:.20/a-3/b=1-----------(2)#

#:.(1) xx 4#

#:.20/a+8/b=12#------(3)

#:.(2)-(3)#

#:.-11/b=-11/1#

#:.b/-11=1/-11#

multiply each side by eleven.

#:.b=-11/-11#

#:.b=1#

change b=1 in (1)

#:.5/a+2=3#

#:.5/a=3-2#

#:.5/a=1#

#:.a/5=1#

multiply each side by five.

#:.a=5#

#:.a=x+y=5#-----(4)

#:.b=x-y=1#------(5)

#:.(4)+(5)#

#:.2x=6#

#:.x=6/2#

#:.x=3#

#:.3+y=5#

#:.y=5-3# in (4)

#:.y=2#

Genesis Allen · Answer

undefined To solve a system of equations using the elimination method, follow these steps:

1. Write down the system of equations.
2. Multiply one or both equations by a constant if necessary to make the coefficients of one of the variables in the two equations equal in magnitude but opposite in sign.
3. Add or subtract the equations to eliminate one of the variables.
4. Solve the resulting equation for the remaining variable.
5. Substitute the value found for the remaining variable into one of the original equations to find the value of the other variable.
6. Check the solution by substituting both values into both original equations to ensure they satisfy both equations.

Repeat these steps as needed until you have found the values of all variables.

Remember, the key idea in the elimination method is to add or subtract the equations in such a way that one of the variables is eliminated, allowing you to solve for the other variable.