A rose costs $2.5 and a carnation costs $3.
Step-by-step explanation:
Let,
Roses = x
Carnations = y
According to given statement;
8x+10y=50 Eqn 1
6x+12y=51 Eqn 2
Multiplying Eqn 1 by 6;
[tex]6(8x+10y=50)\\48x+60y=300\ \ \ Eqn\ 3\\[/tex]
Multiplying Eqn 2 by 8;
[tex]8(6x+12y=51)\\48x+96y=408\ \ \ Eqn 4[/tex]
Subtracting Eqn 3 from Eqn 4;
[tex](48x+96y)-(48x+60y)=408-300\\48x+96y-48x-60y=108\\36y=108[/tex]
Dividing both sides by 36;
[tex]\frac{36y}{36}=\frac{108}{36}\\y=3[/tex]
Putting y=3 in Eqn 3;
[tex]48x+60(3)=300\\48x+180=300\\48x=300-180\\48x=120[/tex]
Dividing both sides by 48;
[tex]\frac{48x}{48}=\frac{120}{48}\\x=2.5[/tex]
A rose costs $2.5 and a carnation costs $3.
Keywords: linear equations, subtraction
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