solve 4y-3x+16=0; 2x-3y=0 by substitution , elimination, and cross multiplication method.
Plz solve by all the three methods and also it on notebook.
plz plz solve
don't spam
Answers
Step-by-step explanation:
Given :-
4y-3x+16=0;
2x-3y=0
To find :-
Solve the equations by substitution , elimination, and cross multiplication methods ?
Solution :-
Substitution Method:-
Given pair of linear equations in two variables are
4y-3x+16 = 0 ----------(1)
and
2x-3y = 0
=> 2x = 0+3y
=> 2x = 3y
=>x = 3y/2 --------------(2)
On substituting the value of x in (1) then
=> 4y -3(3y/2)+16 = 0
=> 4y-(9y/2)+16 = 0
=> (8y-9y+32)/2 = 0
=>(32-y)/2 = 0
=> 32-y= 0×2
=> 32-y = 0
=> y = 32
On Substituting the value of y in (2) then
=> x = 3(32)/2
=> x = 3×16
=> x = 48
Therefore, x = 48 and y = 32
Elimination Method :-
Given pair of linear equations in two variables are
4y-3x+16 = 0
=> -3x+4y +16 = 0
=> -3x+4y = -16
=> 3x-4y = 16
On multiplying with 2 then
=> 6x-8y = 32----------(1)
and
2x-3y = 0
On multiplying with 3 then
=> 6x-9y = 0 ----------(2)
On subtracting (2) from (1) then
6x-8y = 32
6x-9y = 0
(-) (+)
_________
0+y = 32
_________
y = 32
Now , (2) becomes
=> 6x-9(32) = 0
=> 6x - 288 = 0
=> 6x = 288
=> x = 288/6
=> x = 48
Therefore, x = 48 and y = 32
Cross Multiplication Method :-
Given pair of linear equations in two variables are
4y-3x+16 = 0
=> -3x+4y +16 = 0
On comparing with a1x+b1y+c1 = 0 then
a1 = -3
b1 = 4
c1 = 16
and
2x-3y = 0
On comparing with a2x+b2y+c2 = 0 then
a2 = 2
b2 = -3
c2 = 0
Now,
x = (b1c2-b2c1)/(a1b2-a2b1)
=> x = [(4)(0)-(-3)(16)]/[(-3)(-3)-(2)(4)]
=> x = (0+48)/(9-8)
=> x = 48/1
=> x = 48
and
y =(c1a2-c2a1)/( a1b2-a2b1)
=> y = [(16)(2)-(0)(-3)]/[(-3)(-3)-(2)(4)]
=> y =(32-0/(9-8)
=> y = 32/1
=> y = 32
Therefore, x = 48 and y = 32
Answer :-
The solution for the given problem is (48,32)
Used Methods:-
- Substitution Method
- Elimination Method
- Cross Multiplication Method
Used formulae:-
- x = (b1c2-b2c1)/(a1b2-a2b1)
- y =(c1a2-c2a1)/( a1b2-a2b1)