SOLUTION: How do i solve:-4x-y=-6 4x+3y=18

Question 27571: How do i solve:-4x-y=-6
4x+3y=18
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
How do i solve:-4x-y=-6
4x+3y=18
There are two equations in the two unknowns x and y
The easiest and the closest to heart method is to express one variable in terms of the other from one equation and substituting it in the other equation.
What is the idea?
The given two differen equations in x and y are reduced to a single equation in one variable which we have learnt in the LKG class of Algebra!
Which variable that single varible equation result in?
For that we have to study the given equations.
For instance now in the above equations it is easy to talk about y in terms of x as the coefficient for y is (-1)
[1 or -1 is numeriacally 1 and easy to handle]
-4x-y = -6 ----(1)
4x+3y= 18 ----(2)
From (1), we have y =6-4x ----* (taking -y to the other side and bringing -6 to this side and entire sides switching does not ask for sign change. What I mean is initially you get 6-4x = y which we write y= 6-4x)
Putting * in (2), we get
4x + 3(6-4x) = 18
4x + (18 - 24x) = 18 ----(**)
4x + (-24x + 18) =18 (using additive commutativity)
(4x-24x) + 18 = 18 (using additive associativity)
-20x =0 (adding -18 to both the sides and using a+(-a) = 0 in other words canelling 18 additively on both the sides)
Now (-20) X (what ) = 0
The answer is zero.
So we have x= 0
And putting x= 0 in (*), y = 6-4x becomes y = 6- 4X0= 6
We have x = 0 and y= 6
Verification:
Put x=0 and y=6 in (2) and check
LHS = 4x + 3y = 4 X 0 + 3 X 6 = 0 + 18 = 18 = RHS which is correct.
One doubt? Should you check for the correctness of values always in the second equation?
No, not necessarily. If your value for the single variable has resulted because of using (1) in (2) then use (2) for checking and vice versa. That is if you have utilised (2) for finding the single variable and then use (1) for checking the values.If you are in confusion just check both the equations. No harm
One observation:
In the above two equations the coefficients of x are numerically the same and are of opposite signs. This is an ideal situatio.Why?
Beacuse you may just add the two equations(the left with the left and the right with the right)get rid of the x variable, solve for y and use y in one of the equations find x and then check for x and y in the other equation. Cool. Is n't it?
One other observation.
Rules are there for us to follow meticulously when the intution and inspection methods do not work.Otherwise intuition and inspection which are nothing but common sense always score in problem solving.
In the above problem the occurence of 18 on both the sides is a bonus factor!
So mentally cancel and quickly write down 4x - 24x = 0 that is -20x = 0 which means x= 0(of course you can resort to division by (-20) on both the sides and then 0/(anything) = 0 and so x=0 is fine)
Note: So we have understood that if a variable has coefficients equal in magnitude and opposite in sign in the two equations, the equations are addable to get rid of that variable resulting in a single equation variable, as an immediate flash we should ask ourselves the question what if the coefficients of a variable are equal in magnitude and also in sign?
Fine. How to get rid of that variable? Common sense tells us that two equal things vanish when we take away one from the other. Right? So we have to take away one equation from the other to achieve the getting rid of the variable part.
Let me give you two illustrative examples.
Example (1) 3x +5y = 11 and 7x - 5y = 9
I shall now present only those steps which you ought to write in the test
3x + 5y = 11 ----(1)
7x - 5y = 9 ----(2)
(1) + (2) implies
(3x+7x) = (11+9) That is 10x = 20 giving x = 2
x=2 in (1) gives
6 +5y = 11
5y = 11-6
5y = 5 giving y = 1
Answer: x = 2 and y = 1
[we have used (1) for finding the next variable.So check the correctness of x and y in (2)]
Verification: Putting x = 2 and y = 1 in (2)
LHS = 7x - 5y = 7 X 2 - 5 X 1 = 14 - 5 = 9 = RHS which is correct
Example (2) 11x + 5y = 53 and 3x + 5y = 29
11x + 5y = 53 ----(1)
3x + 5y = 29 ----(2)
(1) - (2) implies
(11x - 3x) = (53-29)
8x = 24 giving x = 3
Put x = 3 in (2)
Then 3 X 3 + 5y = 29
9 + 5y = 29
5y = 29 - 9
5y = 20
y = 4
Answer: x = 3 and y = 4
Verification: Put x = 3 and y = 4 in (1)
LHS= 11x + 5y = 11 X 3 + 5 X 4 = 33 + 20 = 53 = RHS which is correct
One Question? What if the coefficients of both the variables are not equal even numerically in the given equations? How to use the above method.
The answer is very simple. If they are not equal, make them equal. How?
Just watch this example.
Example (3) x + 3y = 5 and 13x - 7y = -4
x + 3y = 5 -----(1)
13x - 7y = -4 ----- (2)
(1) X 13 gives
13x + 39 y = 65 ----(I)
13x - 7y = -4 ----- (2) (writing (2) underneath (I) to proceed in the usual manner)
(I) - (2) implies
[39y - (-7y)] = 65 - (-4)
39y + 7y = 65 + 4
46y = 69
y = 69/46 = 3X23/2X23 = 3/2
Put y = 3/2 in (1)
x + 3x3/2 = 5
x + 9/2 = 5
x = 5 - 9/2 = (10-9)/2 = 1/2
Answer: x = 1/2 and y = 3/2
Verification: Put x = 1/2 and y = 3/2 in (2)
LHS = 13x - 7y = 13x(1/2) - 7x(3/2) = 13/2 - 21/2 = -8/2 = -4 = RHS which is correct.
One other question? What if one of the coefficients is not 1 ?
Let us work out an example for this too!
Example 4) 5x + 9y = -4 -----(1) X 2
14x -2y = 16 -----(2) X 9
[The choice is to make the coefficients of y the same as addition of equations is preferable to subtraction. Observe the numerical coefficients: 9 and 2
To equalise, we find the lcm of 9 and 2 which is 18 and hence we multiply (1) through out by 2 and (2) through out by 9]
We get 10x + 18y = -8 ----(3)
and 126x - 18y = 144 ----(4)
(3) + (4) implies
136x = 136 giving x = 1
And x = 1 in (1) implies 5 + 9y = -4
9y = -4-5
9y = -9 giving y = -1
Answer: x=1 and y = -1
which are correct by inspection
Note: There is a method called crossmultiplication method too to solve simultaneous equations in two varibles. If you wish to solve it by that method you may ask for it.

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