  Question: Solve the substitution method, elimination method, cross multiplication method. 5x+2y+13=0, 7x-5y+26=0

Posted by: harjot on 08.06.2021

P
Pavan P. from Bangalore
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Solving in the Substution Method we get:

5x+2y+13=0

∴5x+2y=-13

and 7x-5y+26=0

∴7x-5y=-26 Suppose: Consider; 5x+2y=-13→(1) and 7x-5y=-26→(2) Taking equation (1), we have 5x+2y=-13 ⇒5x=-2y-13 ⇒x=-2y-135→(3) Putting x=-2y-135 in equation (2), we get  7x-5y=-26 7(-2y-135)-5y=-26 ⇒-14y-91-25y=-130 ⇒-39y-91=-130 ⇒-39y=-130+91 ⇒-39y=-39 ⇒y=1→(4) Now, Putting y=1 in equation (3), we get x=-2y-13 x=-2(1)-135 ⇒x=-2-135 ⇒x=-155 ⇒x=-3 ∴x=-3 and y=1.

Solving in the Elimination Method we get:

5x+2y+13=0 ∴5x+2y=-13 and 7x-5y+26=0 ∴7x-5y=-26 5x+2y=-13→(1) 7x-5y=-26→(2) equation(1)×5⇒25x+10y=-65 equation(2)×2⇒14x-10y=-52  Adding ⇒39x=-117 ⇒x=-11739 ⇒x=-3 Putting x=-3 in equation (1), we have 5(-3)+2y=-13 ⇒2y=-13+15 ⇒2y=2 ⇒y=1 ∴x=-3 and y=1.

Solving in the Cross Multiplication Method we get:

5x+2y+13=0 ∴5x+2y=-13 and 7x-5y+26=0 ∴7x-5y=-26 5x+2y+13=0→(1) 7x-5y+26=0→(2) Here, a1=5,b1=2,c1=13 a2=7,b2=-5,c2=26  x=(b1⋅c2-b2⋅c1)/(a1⋅b2-a2⋅b1) =(2)(26)-(-5)(13)(5)(-5)-(7)(2) =(52)-(-65)(-25)-(14) =117-39 x=-3 y=(c1⋅a2-c2⋅a1)/(a1⋅b2-a2⋅b1) =(13)(7)-(26)(5)(5)(-5)-(7)(2) =(91)-(130)(-25)-(14) =-39-39 y=1 ∴x=-3 and y=1.

V
Vikas K. from Lucknow
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X=-3 and Y=1 P
Priya N. from Mumbai
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5x+2y+13=0 7x-5y+26=0 5x+2y= -13 .......... (1) 7x-5y= -26.............(2) Multiplying equation (1) by 5 25x+10y=-65 ......... (3) Multiplying equ(2) by 2 14x-10y= -52......... (4) Adding equ (3) and (4) 25x+10y= -65 +14x-10y=-52 We get, 39x= -117 X= -117/39 X = -3 Place "x=-3" in equ (1) 5(-3) +2y = -13 -15+2y= -13 2y= -13+15 2y= 2 Y= 2/2 Y= 1 x= -3 and y=1 is the solution
S
Shakuntla G. from Noida
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7x-5y+26=0.............1

5x+2y+13=0................2

multiply equation 1 by 2 and equation 2 by 5 and add the resultant equations:

14x-10y+52

25x+10y+65

after adding both result will be 39x+117=0

now the value of x will be -117/39 =3

now put value of x=3 in equation 1 anf yoy will get value of y

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