6x-5y=21 and 2x+3y=19. find x and y. please teach me how to solve this kind of equation.?

Question

I do know how to solve when an equation is like this :6x-3y=21 and2x+3y=19 where the +3y and -3y cancelled out each other and become zero.
What about an equation like this:6x-5y=21 and 2x+3y=19. find x and y.
please teach me how to solve this kind of equation

Answer 1

Multiply constants to one or both equations until either x or y can cancel out

6x - 5y = 21
2x + 3y = 19

6x - 5y = 21
-6x - 9y = -57 (2nd equation by -3)
_____________
-14y = -36
y=36/14 or 18/7
Then just substitute y to one equation and solve for x.

Answer 2

In these kinds of equations, there are many ways to find x & y. One simple method is to solve using your method -
First make the value of either x or y equal in both equations by multiplying the whole equation by some number.i.e. 6x -5y=21
and 3{2x+3y=19} will make the value of x equal in both cases. The new equations would be 6x -5y=21 and 6x+9y=57 which you can easily solve by subtracting the former from the latter. This process is called elimination method.

Answer 3

6x -5y=21 .............. Eqn (1)
2x+3y=19 ............... Eqn (2)

First of all mark that eqn (1) has 6 as coefficient of x, while in eqn (2) coefficient of x is 2. Try to change this 2 to 6 so that x has the same ( ie 6) as coefficient in both the eqns. Naturally you will have to multiply (2x) by 3 to get 6x for the second eqn. But whenever you multiply or divide the left hand side (LHS) of any eqn, you must do the same to RHS too. So that we have -


6x - 5y = 21 .............. Eqn (1) and the changed eqn (2) is
6x + 9y = 57 ................ Eqn (3)
___________
0 - 14y = - 36 (On subtracting the two eqns)
ie y = 36/14 = 18/7
Now substitute this value in eqn 2

ie 2 x + 3 (18/7) = 19
ie x = 79/14

Hence x = 79/14 and y = 18/7 ............... Answer

Answer 4

6x - 5y = 21
2x + 3y = 19

answer
let say
6x - 5y = 21 is equation 1, and
2x + 3y = 19 is equation 2
if we x3 the equation 2, it will be
3(2x + 3y) = 3(19)
we get
6x + 9y = 57 and let say it is equation 3
if equation 3 minus equation 1,
6x + 9y = 57
- (6x - 5y = 21)
--------------------
0 +14y = 36
so, 14y = 36
y = 36/14
y = 18/7
then we replace the value of y = 18/7 into the equation 1
6x - 5(18/7) = 21
6x - 90/7 = 21
6x = 90/7 + 21
6x = (90 + 147)/7
6x = 237/7
x = 237/(7 x 6)
x = 79/(7 x 2)
x = 79/14

hope it will help you

Answer 5

6X-5Y=21
2X+3Y=19
These are called simultaneous equations.we solve them by elimination; substitution; graphically or matrix method. i'll use elimination method.
i'll eliminate the x.
i.e 6x-5y=21
3(2x+3y=19)
6x-5y=21
6x+9y=57 ----minus equation 2 from wquation 1
we have now;
14y=36
y=18/7
from equation 2; 2x=19-3(18/7) x=38/7

Answer 6

just get y in one of the equations....for a definite explanation i'll use 6x-5y=21, where y is,

y=(21-6x)/5

substitute y in terms of x in the second equation to solve for x.

2x + 3/5(21-6x) = 19
2x + (63-18x)/5 = 19
10x + 63 - 18x = 95
-8x = 95-63
x = -32/8
x = -4 (answer)

substitute the value of x in the first equation.

6x - 5y = 21
6(-4) - 5y = 21
-5y = 21+24
y = -45/5
y = -8 (answer)

i hope this helps a lot......

Answer 7

When you have 2 unknowns in these equations, it's important to make the unknowns to 1.
eg. (1) 6x-5y=21 and (2) 2x+3y=19
from (1), let's make x as the subject formula
6x=21-5y
x=(21-5y)/6 (let's name this eqn 3)
Substitute eqn (3) into (2)
2[(21-5y)/6] + 3y =19
Now, you only have one unknown that is y. Solve y and substitute it into eqn (1) to get x. Voila!

Answer 8

Okay Here is the way to do it using that method. Your tryng to find the LCM .
SO lets do it for x.
FInd the LCM of 6 and 2. It would be 6.
SO, lets multiply the 2nd equation by -3.
That gives you -6x-9y=-57.
SO adding both equaTIONS, CANCELING OUT THE X, that gives you -14y=-36, or y=-36/-14 = 36/14.
And x=(21+((36/14)*5))/6.

Answer 9

6x-3y=21
adding
2x+3y=19
-----------------
8x=40
x=40/8
x=5
substituting in one first equation.
6x-5y=21
x=5
6(5)-5y=21
30-5y=21
30-21=5y
9=5y
y=9/5


here the main aim is to cancel out one variable.since 3y is of different signs we can cancel it out by adding the two equations.find the value of one variable then substitute in any one equation for other variable's value.

6x-5y=21(EQ1)
2x+3y=19(EQ2)
the simplest way to solve this is to make the coefficents of y the same as they will cancel out on adding.for this multiply coeff of y in eq1 with eq2 and coeff of eq2 with eq1.

(6x-5y=21)*3
(2x+3y=19)*5
--------------------
18x-15y=63
adding
10x+15y=95
--------------
28x=158
x=158/28
x=79/14
substitute in eq 1
6(79/14)-5y=21
(474/14)-21 =5y
(474-21*5)/14=5y
474-105=14*5y
369=70y
y=369/70

Answer 10

You can use cross rule to solve this eqn. you start with coefficient of y and end with the same.
-5 -21 6 -5
3 -19 2 3

x = y = 1
----------- ----------- ------------------
3x-21-(-5)x(-19) -21x2-(-19)x6 6x3-2x(-5)
after that you choose the 1st and 3rd , 2nd and 3rd.