Idont know how to solve systems by substitution

Solve the second equation for x ( x = -13 - 3y) and then substitute the value into first equation

Solve the first equation for y( y = 2x - 9 ) and then substitute that value into second equation

solve either equation for 3y ( 3y = -13 - x  or 3y = 6x - 27) and then substitute that value into other equation

Solution:

Given system of equations are:

6x - 3y = 27  ------- eqn 1

x + 3y = -13 -------- eqn 2

We have to solve the above system of equations by substitution method

Solve the second equation for x ( x = -13 - 3y) and then substitute the value into first equation

Solve the eqn 2 for variable "x"

From eqn 2,

x + 3y = -13

x = -3y - 13 ------- eqn 3

Substitute eqn 3 in eqn 1

6(-3y - 13) - 3y = 27

-18y - 78 - 3y = 27

Combine the like terms

-18y - 3y - 78 = 27

-21y = 27 + 78

-21y = 105

y = -5

Substitute y =-5 in eqn 2

x + 3(-5) = -13

x - 15 = -13

x = -13 + 15

x = 2

Thus the solution is (x, y) = (2, -5)

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3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.