it says 'solve the following systems of equations. solve one using substitution and solve the other using elimination' the problems are x - 3y = 16 and 3x - 2y = 13 i don't know how to solve these please help

Step-by-step explanation:

you gave me only one system of 2 equations. not 2 systems or more.

maybe you missed something.

I can solve this one system in two ways, though.

substitution :

x - 3y = 16

3x - 2y = 13

the first equation gives us

x = 3y + 16

that we can use now in the second equation

3×(3y+16) - 2y = 13

9y + 48 - 2y = 13

7y + 48 = 13

7y = -35

y = -5

x = 3y + 16 = 3×-5 + 16 = -15 + 16 = 1

that's it. that is all the secret of substitution. you use one equation to express one variable by the other. and then you use that in the second equation to have suddenly only one equation with one variable that you can solve. and then from that result you calculate back the other "substituted" variable.

elimination :

x - 3y = 16

3x - 2y = 13

we multiply one equation completely by a factor and then add both equations. of course we should pick a factor that will eliminate one variable, when we add the equations.

e.g.

multiply the first equation by -3

-3x + 9y = -48

3x - 2y = 13

0 7y = -35

y = -5

now we pick any one of the equations to use that result and calculate the other variable

e.g. the second equation

3x - 2×-5 = 13

3x + 10 = 13

3x = 3

x = 1

done.

as you can see : of course, both results are identical.

use the described principles to solve any other systems of equations you might have missed here.

g

step-by-step explanation:

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb