1) <span>B)3x+6y=22
6x+12y=44
because the equations are equivalent to each other
2*(3x+6y=22) -----> 6x+12y=44
2) </span><span>5x+10y=15 should be multiplied by -2, to get 5x*(-2)=-10x, and x then can be eliminated
</span><span>A)-2</span>
Answer:
Kindly check explanation
Step-by-step explanation:
Given that :
Total flowers sold = 200
Flowers are :
Carnations, Roses, and Gerber daisies
Carnations = c
Roses = r
Daisies = d
20 fewer roses than daisies ; r = d - 20
Total sales = $453
c = $1.50 each ; r = 3.75 each ; d = 2.25 each
c + d + d - 20 = 200
c + 2d = 220 - - (1)
1.50c + (d - 20)(3.75) + 2.25d = 453
1.50c + 3.75d - 75 + 2.25d = 453
1.50c + 6d = 528 - - - (2)
From (1)
c = 220 - 2d
1.5(220 - 2d) + 6d = 528
330 - 3d + 6d = 528
330 + 3d = 528
3d = 528 - 330
d = 198/3
d = 66
66 × 2.25 = $148.5
c = 220 - 2(66)
c = 88
88 * 1.50 = $132
r = d - 20
r = 66 - 20
r = 46
46 × $3.75 = $172.5
1.
(x/2) - 7 = 11
2.
2x + 7 = 27
3.
2x - 5 = 25
Hope this helped!! (:
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;
4y − x + 4 = 0 ;
________________________________________________________
"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ;
→ 4y − x + 4 = 0 ;
Subtract "4" from each side of the equation ;
→ 4y − x + 4 − 4 = 0 − 4 ;
→ 4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________
y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;
4y − x = -4 ; ===> Refer to this as "Equation 2" ;
________________________________________________________
Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;
→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;
to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :
→ " 4y − x = - 4 " ;
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;
→ as follows:
_________________________________________________
→ " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________
a(b + c) = ab + ac ; AND:
a(b − c) = ab <span>− ac .
_________________________________________________
As such:
We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:
→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:
→ " -16x + 64 − x = - 4 " ;
Note: " - 16x − x = -16x − 1x = -17x " ;
→ " -17x + 64 = - 4 " ; Solve for "x" ;
Subtract "64" from EACH SIDE of the equation:
→ " -17x + 64 − 64 = - 4 − 64 " ;
to get:
→ " -17x = -68 " ;
Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -17x / -17 = -68/ -17 ;
to get:
→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;
which is: " y = -4x + 16" ; to solve for "y".
______________________________________
→ y = -4(4) + 16 ;
= -16 + 16 ;
→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;
→ Let us check;
→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;
→ Consider the first equation given in our problem, as originally written in the system of equations:
→ " y = - 4x + 16 " ;
→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?
→ "0 = ? -4(4) + 16 " ?? ; → "0 = ? -16 + 16 ?? " ; → Yes! ;
{Actually, that is how we obtained our value for "y" initially.}.
→ Now, let us check the other equation given—as originally written in this very question:
→ " 4y − x + 4 = ?? 0 ??? " ;
→ Let us "plug in" our obtained values into the equation;
{that is: "4" for the "x-value" ; & "0" for the "y-value" ;
→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.
→ " 4(0) − 4 + 4 = ? 0 ?? " ;
→ " 0 − 4 + 4 = ? 0 ?? " ;
→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________
