Two white roses and one yellow rose cost 5 pounds. Two white roses and three red roses cost 10.50 pounds. Three yellow roses and two red roses cost 11 pounds. What is the total cost of one red rose, one white rose and one yellow rose?

Step-by-step solution

1. 1

   Name the unknowns and write the three equations. Let $w$, $y$ and $r$ be the price of one white, one yellow and one red rose:

   $2w+y=5,\qquad 2w+3r=10.50,\qquad 3y+2r=11.$

   Each equation involves only two of the three colours, which is the hint that substitution will be cleaner than elimination here.

2. 2

   Express two variables in terms of the third. The first and second equations each contain $2w$, so solving both for the other variable keeps $w$ as the single unknown:

   $y=5-2w,\qquad r=\frac{10.50-2w}{3}=3.50-\frac{2}{3}w.$

3. 3

   Substitute into the equation that has no $w$ in it. The third equation is the only one linking $y$ and $r$, so it becomes an equation in $w$ alone:

   $3(5-2w)+2\left(3.50-\tfrac{2}{3}w\right)=11\ \Longrightarrow\ 15-6w+7-\tfrac{4}{3}w=11.$

4. 4

   Solve for the white rose price. Combining the $w$ terms with $6=\tfrac{18}{3}$:

   $22-\frac{22}{3}w=11\ \Longrightarrow\ \frac{22}{3}w=11\ \Longrightarrow\ w=1.50.$

5. 5

   Back-substitute and add. From $y=5-2(1.50)=2.00$ and $2(1.50)+3r=10.50\Rightarrow r=2.50$:

   $w+y+r=1.50+2.00+2.50=\pounds 6.00.$

6. 6

   Verify all three original conditions. $2(1.50)+2.00=5$; $2(1.50)+3(2.50)=10.50$; $3(2.00)+2(2.50)=11$. All three hold exactly, so the prices - and the total of $\pounds 6.00$ - are correct.