If 0.5x+0.7y = 5.8, 2.1 x - 1.4y = -14.7 and px= 5y-36, then the value of p is
Please tell the correct answer with solution
Answers
Step-by-step explanation:
Given:-
0.5x+0.7y = 5.8,
2.1 x - 1.4y = -14.7
and px= 5y-36
To find:-
Find the value of p ?
Solution:-
Given equations :
0.5x+0.7y = 5.8 ---------(1)
On multiplying with 10 both sides then
=>10×(0.5x+0.7y )= 5.8×10
5x +7y = 58
On multiplying with 2 then
10x+14y = 116 ------------(2)
2.1 x - 1.4y = -14.7-------(3)
On multiplying with 10 both sides then
=> 10×(2.1 x - 1.4y) = -14.7×10
21x-14y = -147-----------(4)
On adding (2)&(4) then
10x+14y = 116
21x-14y = -147
(+)
___________
31x +0 = - 31
___________
=> 31x = -31
=> x = -31/31
=> x = -1
On Substituting the value of x in (2)
10x+14y = 116
=> 10(-1)+14y = 116
=> -10+14y = 116
=> 14y = 116+10
=> 14y = 126
=> y = 126/14
=> y = 9
Therefore, x = -1 and y = 9
and given another equation is px= 5y-36
On Substituting these values in the above equation then
=> p(-1) = 5(9)-36
=> -p = 45-36
=> -p = 9
=> p = -9
Answer:-
The value of p for the given problem is -9
Check:-
If x = -1 and y = 9 then 0.5x+0.7y
=> 0.5(-1)+0.7(9)
=> -0.5+6.3
=> 5.8
=> RHS
LHS = RHS is true for x = -1 and y = 9
If x = -1 and y = 9 then 2.1 x - 1.4y
=> 2.1(-1)-1.4(9)
=> -2.1 - 12.6
=> -14.7
=> RHS
LHS = RHS is true for x = -1 and y = 9
and If p = -9 then
px = (-9)(-1)
px = 9
and
5y-36
=> 5(9)-36
=> 45-36
=>9
LHS = RHS is true for x = -1 ,y = 9 and p = -9
Used Method :-
- Method of Elimination.