Answer:
[tex]C=\frac{17}{7}, D=-\frac{4}{7}[/tex]
Explanation:
The system of equation is the following:
[tex]3C+4D=5\\2C+5D=2[/tex]
We can solve the system by expliciting C from the second equation. We get:
[tex]2C+5D=2\\2C=2-5D\\C=\frac{2-5D}{2}=1-\frac{5}{2}D[/tex]
And if we know substitute C into the first equation, we find
[tex]3C+4D=5\\3(1-\frac{5}{2}D)+4D=5\\3-\frac{15}{2}D+4D=5\\3-\frac{15}{2}D+\frac{8}{2}D=5\\3-\frac{7}{2}D=5\\-\frac{7}{2}D=2\\-7D=4\\D=-\frac{4}{7}[/tex]
And by substituting D into the second equation, we find
[tex]C=1-\frac{5}{2}D=1-\frac{5}{2}(-\frac{4}{7})=1+\frac{10}{7}\\C=\frac{7}{7}+\frac{10}{7}=\frac{17}{7}[/tex]
