I buy a number of cheap calculators for a total of $240, mark each of them up by$2 and sell all but four of them for a total of $240. How many calculators did I buy? Okay, so this is what I have done so far: Let y represent the cost per calculator, and let x represent the number of calculator. Therefore, xy = 240 And with the addition of the$2.00, minus the four pencils:
(x-4)(y+2) = 240
xy+2x-4y-8 = 240

and now im stumped. what do i do?
2. Let $\displaystyle x$ be the no. of calculators. Cost per calculator=\displaystyle \frac {240}{x}\displaystyle (x-4)\left(\frac {240}{x}+2\right)=240\displaystyle \Rightarrow (x-4)\left(\frac{240+2x}{x}\right)=240\displaystyle \Rightarrow (x-4)(240+2x)=240x\displaystyle \Rightarrow 240x+2x^2-960-8x=240x\displaystyle \Rightarrow 2x^2-8x-960=0\displaystyle \Rightarrow x^2-4x-480=0\displaystyle \Rightarrow (x+20)(x-24)=0\displaystyle \Rightarrow x=-20$(Impossible) or$\displaystyle x=24$So$\displaystyle 24$is the required answer 3. Originally Posted by foreverbrokenpromises I buy a number of cheap calculators for a total of$240, mark each of them up by $2 and sell all but four of them for a total of$240. How many calculators did I buy?
And with the addition of the $2.00, minus the four pencils: (x-4)(y+2) = 240 xy+2x-4y-8 = 240 and now im stumped. what do i do? please explain + thanks in advance! Now you have to substitute$\displaystyle xy=240\displaystyle 240+2x-4y-8=240\displaystyle \Rightarrow 2x-4y=8\displaystyle \Rightarrow x-2y=4\displaystyle \Rightarrow x=4+2y$Now$\displaystyle xy=240\displaystyle \Rightarrow (4+2y)y=240\displaystyle \Rightarrow 2y^2+4y-240=0\displaystyle \Rightarrow y^2+2y-120=0\displaystyle \Rightarrow (y+12)(y-10)=0\displaystyle \Rightarrow y=-12$(not possible ) or$\displaystyle y=10\displaystyle \Rightarrow x=4+2\times10=24\$