Math, asked by goodbrainly, 1 year ago

Solve the system of equations
-x/2 + y/3 = 0

x + 6y = 16

Answers

Answered by TheRuhanikaDhawan
0
multiply first equation by 6


6(-x/2 + y/3) = 6(0) 


-3x + 2y =0 ------------------- 1


x + 6y = 16 


x = 16 - 6y ----------------------2


substitute 2 in 1


-3(16 -6y ) + 2y = 0

-48 y + 18 y + 2y =0

-48 + 20y =0


48/20=y


12/5 =y                                                                               
                                                                               
substiute y in 2


x + 6(12/5) = 16


x = 8/5                                                                            
                                  
Answered by ROYALJATT
0
we will solve the system of equation by subsitution method
equation 1 

6(-x/2 + y/3) = 6(0) { multiplying equation 1}

so we get
 -3x + 2y =0 

equation 2

x + 6y = 16 
{taking value of x for subsitution equation 1}



now subsituting


-3(16 -6y ) + 2y = 0

-48 y + 18 y + 2 y =0

-48 + 20 y =0

y=48/20

y=12/5                                                                                
                                                                               
substiuting y in equation 2


x + 6(12/5) = 16


x = 8/5                                                                           
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