Let's call C to the cups of chocolate chips and S to the cups of sugar. We are told that the cups of sugar are 2 2/3 times the cups of chocolate, then we can formulate the following equation:
![S=2(2)/(3)C](https://img.qammunity.org/2023/formulas/mathematics/college/faxwolidz7mbwrxp193jvcpfrq25oi9gf6.png)
In the case 4 1/3 of sugar is added, we can replace 4 1/3 for S to get:
![4(1)/(3)=2(2)/(3)C](https://img.qammunity.org/2023/formulas/mathematics/college/dfghje0s3ldkphxfeqhyjscym18sldi0sz.png)
By dividing both sides by 2 2/3 we get:
![\begin{gathered} 4(1)/(3)/2(2)/(3)=2(2)/(3)C/2(2)/(3) \\ 4(1)/(3)/2(2)/(3)=C \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r7fow3zrjd8l4pviwwo7q8ipauesrdhsm6.png)
We can rewrite the mixed fractions to get:
![\begin{gathered} (4*3+1)/(3)/(2*3+2)/(3)=C \\ (12+1)/(3)/(6+2)/(3)=C \\ (13)/(3)/(8)/(3)=C \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9uznel1ze8wjl2v8c34obm4ugbf5zh1kwn.png)
By changing the division symbol to a multiplication symbol and flipping the 8/3, we get:
![\begin{gathered} (13)/(3)*(3)/(8)=C \\ (13)/(8)=C \\ (8+5)/(8)=C \\ (8)/(8)+(5)/(8)=C \\ 1+(5)/(8)=C \\ 1(5)/(8)=C \\ C=1(5)/(8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8o2ryik2zfvw2rj20187091b20r1te314v.png)
Then, 1 5/8 cups of chocolate chips are needed