Question

The scouts sold small and large boxes of cookies as a fund raiser. One scout sold 7 small boxes and 12 large boxes for $54.00. Another scout sold 5 small boxes and 10 large boxes for $60.00. The scout leader wrote a system of equations to represent the their sales. 7x+12y=54 5x+10y=60 Which constants can each equation be multiplied by so that one variable is eliminated when the equations are added?
Check all that apply.
A) The first equation can be multiplied by 5 and the second equation by –6 to eliminate the y.
B) The first equation can be multiplied by –5 and the second equation by 6 to eliminate the y.
C) The first equation can be multiplied by –5 and the second equation by 7 to eliminate the x.
D) The first equation can be multiplied by 5 and the second equation by –7 to eliminate the x.
E) The first equation can be multiplied by –5 and the second equation by 10 to eliminate the x.

Answer 1

Answer:

Step-by-step explanation:

A) When the first equation is multiplied by 5 and the second equation by –6 ,

the equations become,

35x + 60y=54

-30x - 60y=60

Hence we can eliminate y by adding the equation.

B)  When the first equation is multiplied by -5 and the second equation by 6 ,

the equations become,

-35x - 60y=54

30x + 60y=60

Hence we can eliminate y by adding the equation.

C)  When the first equation is multiplied by -5 and the second equation by 7,

the equations become,

-35x  - 60y=54

35x - 70y=60

Hence we can eliminate x by adding the equation.

D) When the first equation is multiplied by 5 and the second equation by -7,

the equations become,

35x  + 60y=54

-35x - 70y=60

Hence we can eliminate x by adding the equation.

E) When the first equation is multiplied by -5 and the second equation by 10,

the equations become,

-35x  - 60y=54

50x - 100y=60

Hence we can not eliminate x by adding the equation.