MATH SOLVE

3 months ago

Q:
# In order to solve the system of equations (x-3y=2, 2x+y=11 using the elimination method, which of the following steps could be used? A. Multiply the first equation by 2 and then add that result to the second equationB. Multiply the first equation by 3 and then add that result to the second equationC. Multiply the second equation by 2 and then add that result to the first equationD. Multiply the second equation by 3 and then add that result to the first equation

Accepted Solution

A:

The correct answers are:

__________________________________________________________

[C]: "Multiply the second equation by 2 and then add that result to the first equation" ; AND:

__________________________________________________________

[D]: "Multiply the second equation by 3 and then add that result to the first equation" .

__________________________________________________________

Explanation:

__________________________________________________________

Given:

__________________________________________________________

The first equation: " x − y = 2 " ;

The second equation: " 2x + y = 11 " ;

_____________________________________________________

Consider choice [A]: "Multiply the first equation by 2 and then add that result to the second equation" .

→ Multiply the first equation by "2" :

2 * {x − 3y = 2} ;

→ 2x − 6y = 4 ;

Then, add this to the second equation:

2x − 6y = 4

+ 2x + y = 11

_________________

4x − 5y = 15 ; → RULE OUT "Choice [A]" .

________________________________

Consider choice [B]: "Multiply the first equation by 3 and then add that result to the second equation" ;

→ Multiply the first equation by "3" :

3 * {x − 3y = 2} ;

→ 3x − 9y = 6 ;

Then, add this to the second equation:

3x − 9y = 6

+ 2x + y = 11

_________________

5x − 8y = 17 ; → RULE OUT "Choice [B]" .

____________________________________________________.

Consider choice [C]: "Multiply the second equation by 2 and then add that result to the first equation" ;

→ Multiply the SECOND equation by "2" :

2 * (2x + y = 11} ;

→ 4x + 2y = 22 ;

Then, add this to the first equation:

4x + 2y = 22

+ x − 3y = 11

_________________

5x − y = 33 ; → This is a correct answer choice—since we can easily isolate "y" on one side of the equation:

→ 5x − y = 33 ;

↔ -y + 5x = 33 ;

Subtract "5x" from each side of the equation;

→ -y + 5x − 5x = 33 − 5x ;

→ -y = 33 − 5x ;

Multiply each side of the equation by "-1" ;

→ (-1) * (-y) = (-1) * (33 − 5x) ;

→ y = 5x − 33 ;

____________________________________________________

Consider choice [D]: "Multiply the second equation by 3 and then add that result to the first equation" ;

→ Multiply the SECOND equation by "3" :

3 (2x + y = 11)

→ 6x + 3y = 33 ;

Then, add this to the first equation:

6x + 3y = 33

+ x − 3y = 2

___________________

7x = 35 ;

Now, we can solve for "x" ;

7x = 35 ;

Divide EACH SIDE of the equation by "7" ;

to isolate "x" on one side of the equation; & to solve for "x" ;

7x / 7 = 35/7 ;

x = 5 ; → Yes; this answer choice, [D]; is a correct step.

Furthermore, we can take: "y = 5x − 33 " ; from "choice [D]:

and plug in "5" into "x" ; to solve for "y" ;

→ y = 5(5) − 33 = 25 − 33 = -8.

________________________________________________

So, the answer to this system of equations is:

________________________________________________

"x = 5, y = 8 " ; or, write as: " [5, 8] " .

________________________________________________

__________________________________________________________

[C]: "Multiply the second equation by 2 and then add that result to the first equation" ; AND:

__________________________________________________________

[D]: "Multiply the second equation by 3 and then add that result to the first equation" .

__________________________________________________________

Explanation:

__________________________________________________________

Given:

__________________________________________________________

The first equation: " x − y = 2 " ;

The second equation: " 2x + y = 11 " ;

_____________________________________________________

Consider choice [A]: "Multiply the first equation by 2 and then add that result to the second equation" .

→ Multiply the first equation by "2" :

2 * {x − 3y = 2} ;

→ 2x − 6y = 4 ;

Then, add this to the second equation:

2x − 6y = 4

+ 2x + y = 11

_________________

4x − 5y = 15 ; → RULE OUT "Choice [A]" .

________________________________

Consider choice [B]: "Multiply the first equation by 3 and then add that result to the second equation" ;

→ Multiply the first equation by "3" :

3 * {x − 3y = 2} ;

→ 3x − 9y = 6 ;

Then, add this to the second equation:

3x − 9y = 6

+ 2x + y = 11

_________________

5x − 8y = 17 ; → RULE OUT "Choice [B]" .

____________________________________________________.

Consider choice [C]: "Multiply the second equation by 2 and then add that result to the first equation" ;

→ Multiply the SECOND equation by "2" :

2 * (2x + y = 11} ;

→ 4x + 2y = 22 ;

Then, add this to the first equation:

4x + 2y = 22

+ x − 3y = 11

_________________

5x − y = 33 ; → This is a correct answer choice—since we can easily isolate "y" on one side of the equation:

→ 5x − y = 33 ;

↔ -y + 5x = 33 ;

Subtract "5x" from each side of the equation;

→ -y + 5x − 5x = 33 − 5x ;

→ -y = 33 − 5x ;

Multiply each side of the equation by "-1" ;

→ (-1) * (-y) = (-1) * (33 − 5x) ;

→ y = 5x − 33 ;

____________________________________________________

Consider choice [D]: "Multiply the second equation by 3 and then add that result to the first equation" ;

→ Multiply the SECOND equation by "3" :

3 (2x + y = 11)

→ 6x + 3y = 33 ;

Then, add this to the first equation:

6x + 3y = 33

+ x − 3y = 2

___________________

7x = 35 ;

Now, we can solve for "x" ;

7x = 35 ;

Divide EACH SIDE of the equation by "7" ;

to isolate "x" on one side of the equation; & to solve for "x" ;

7x / 7 = 35/7 ;

x = 5 ; → Yes; this answer choice, [D]; is a correct step.

Furthermore, we can take: "y = 5x − 33 " ; from "choice [D]:

and plug in "5" into "x" ; to solve for "y" ;

→ y = 5(5) − 33 = 25 − 33 = -8.

________________________________________________

So, the answer to this system of equations is:

________________________________________________

"x = 5, y = 8 " ; or, write as: " [5, 8] " .

________________________________________________