Question

Please help!
1.Solve the system using substitution.
X+y=8
Y=3x
2.solve the system using substitution.
2x+2y=38
Y=x+3

3.Solve the system using substitution.
Y=8-x
7=2-y

Answer 1

\[X+Y=8 \ .....(1)\\ Y=3X\ .....(2)\\ substitution\ means\ replacing\ one\ variables\ by\ other\ variables\\ putting\ y=3x \ into\ eq\ (1)\ we've\\x+3x=8\\ \implies\ 4x=8\\implies\ x=\frac{ 8 }{ 4 }\\or\ x=2\\ now\ puting\ x=2\ into\ eq\ (2)\ we've\ y=3\times2=6\\therefore\ x=2\ and\ y=6\ are\ simultaneous\ solution\ of \ eq\ (1) and\ (2).\]

Answer 2

@Casadie1004 similarly u can solve 2 and 3. hope so.

Answer 3

for number 2 you can divide each term in equation 1 by 2 first

Answer 4

2)2x+2y=38----------(1)
y=x+3--------------(2)

2x+2(x+3) =38 [substituted y= x+3 in eq (1)]
simplify and get x value

Answer 5

Question 3\[\huge{y=8-x->(1)}\]\[\huge{7=2-y->(2)}\]Substitute the first equation into the second equation to solve for x\[\huge{7=2-(8-x)}\]\[\huge{7=2-8+x}\]\[\huge{7=-6+x}\]\[\huge{x=13}\]Then,substitute the value of x into the first equation to find y\[\huge{y=8-x}\]\[\huge{y=8-(13)}\]\[\huge{y=?}\]solve it @Casadie1004

Answer 6

Y= _5

Answer 7

Y = -5

Answer 8

Is it clesr

Answer 9

thanks all of you!

Answer 10

yw! ^_^
@Casadie1004