MR-368 / 12. problem

Determine the true value of the following expressions:

$p\equiv \frac{3+2x}{3}-\frac{2-3x}{4}=\frac{29}{24};x=0,5$

$q\equiv 3\left(5-y\right)-2\left(y-1\right)=1+3y;y=2$Then determine the true value of the following two terms:

a) $\left(p\vee q\right)\wedge p$

b) $\left(p\wedge q\right)\vee p$

${p}{\equiv }\frac{3+2·0,5}{3}-\frac{2-3·0,5}{4}{=}\frac{29}{4}$

${p}{\equiv }\frac{3+1}{3}-\frac{2-1,5}{4}{=}\frac{29}{4}$
${p}{\equiv }\frac{4}{3}-\frac{0,5}{4}{=}\frac{29}{4}$
${p}{\equiv }\frac{4}{3}-\frac{1}{4}·\frac{1}{2}{=}\frac{29}{4}$
${p}{\equiv }\frac{4}{3}-\frac{1}{8}{=}\frac{29}{4}$

${p}{\equiv }\frac{4}{3}-\frac{1}{8}=\frac{4·8-3·1}{24}=\frac{32-3}{24}=\frac{29}{24}{=}\frac{29}{24}$

${p}{\equiv }\frac{29}{24}{=}\frac{29}{24}⇒p=\top$

${q}{\equiv }3\left(5-y\right)-2\left(y-1\right){=}{1}{+}{3}{y};y=2$

${q}{=}3\left(5-2\right)-2\left(2-1\right){=}{1}{+}{3}{·}{3}$
${q}{=}3·3-2·1{=}{1}{+}{3}{·}{2}$
${q}{=}9-2{=}{1}{+}{6}$



$\left(p\vee q\right)\wedge p=\top$



$\left(p\wedge q\right)\vee p=\top$

Replace the value of x to in the equation and check its correctness