Respuesta :
ok so lets say the first number is x and the second number is y we will use a system of equations to solve.
"Three times one number is twice the second" this equation is 3x=2y
"Twice the first number is 3 more than the second number." this equation would be 2x=y+3
To solve for x, change the second equation to y= 2x-3 by simply subtracting 3 from y to get y by itself.
Plug 2x-3 in for y in the first equation
3x=2(2x-3) solve for x - distribute the 2
3x=4x-6 subtract the 4x from both sides
-x= -6 divide both sides by -1 to get rid of the -x
x=6 now plug this into x in any of the 2 equations and you will find y
2x=y+3
2(6)=y+3
12=y+3
9=y
first number is 6 second number is 9
"Three times one number is twice the second" this equation is 3x=2y
"Twice the first number is 3 more than the second number." this equation would be 2x=y+3
To solve for x, change the second equation to y= 2x-3 by simply subtracting 3 from y to get y by itself.
Plug 2x-3 in for y in the first equation
3x=2(2x-3) solve for x - distribute the 2
3x=4x-6 subtract the 4x from both sides
-x= -6 divide both sides by -1 to get rid of the -x
x=6 now plug this into x in any of the 2 equations and you will find y
2x=y+3
2(6)=y+3
12=y+3
9=y
first number is 6 second number is 9
Let x be the 1st number & y the 2nd
3x=2y (1st equation)
and
2x=y+3 (2nd equation) ===> 2x-3 = y (3rd equation) & y=9
Substitute in 1st equation, y but its value in the 3rd
===>3x= 2(3x-3)===>solve for x=6
3x=2y (1st equation)
and
2x=y+3 (2nd equation) ===> 2x-3 = y (3rd equation) & y=9
Substitute in 1st equation, y but its value in the 3rd
===>3x= 2(3x-3)===>solve for x=6