Question 1199025
Let {{{x}}} be the grams of carbohydrates in a cup of raw lettuce, {{{y}}} for raw asparagus spears, and {{{z}}} for raw tomatoes. Then, we have the system of equations {{{system(x+6y+z=12,x+6y=4.5z, x+z=18y)}}}. Subtracting {{{z}}} from both sides of the first equation, we get {{{x+6y=12-z}}}. Subtracting that from the second equation, we get {{{5.5z=12}}}. Dividing both sides by {{{5.5}}} gives {{{z=12/5.5=highlight(24/11)}}}.
If we subtract {{{18y}}} from both sides of the third equation, we get {{{x-18y+z=0}}}. Subtracting that from the first equation gives {{{24y=12}}}. Dividing both sides by 24 gives {{{y=highlight(1/2)}}}.
We can plug the values that we found for {{{y}}} and {{{z}}} into the first equation to get {{{x+6*1/2+24/11=12}}}. Simplifying, we get {{{x+3+24/11=12}}}. We isolate {{{x}}} to get {{{x=12-3-24/11=9-24/11=99/11-24/11=highlight(75/11)}}}.
In conclusion, the solutions are {{{system(x=75/11,y=1/2,z=24/11)}}}.
Check: {{{75/11+6*1/2+24/11=99/11+3=9+3=12}}}
{{{75/11+6*1/2=75/11+3=75/11+33/11=108/11=4.5*24/11=4.5z}}}.
{{{75/11+24/11=99/11=9=18*1/2=18y}}}.