Solve (t+1)/2 + (t-3)/3 = (2t+5)/6
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Answers
+12+−33=(2+5)6
\frac{t+1}{2}+\frac{t-3}{3}=\frac{(2t+5)}{6}2t+1+3t−3=6(2t+5)
3(+1)6+2(−3)6=(2+5)6
3(+1)6+2(−3)6=(2+5)6
\frac{3(t+1)}{6}+\frac{2(t-3)}{6}=\frac{(2t+5)}{6}63(t+1)+62(t−3)=6(2t+5)
3(+1)+2(−3)6=(2+5)6
3(+1)+2(−3)6=(2+5)6
\frac{{\color{#c92786}{3(t+1)}}+2(t-3)}{6}=\frac{(2t+5)}{6}63(t+1)+2(t−3)=6(2t+5)
3+3+2(−3)6=(2+5)6
3+3+2−66=(2+5)6
\frac{3t+{\color{#c92786}{3}}+2t{\color{#c92786}{-6}}}{6}=\frac{(2t+5)}{6}63t+3+2t−6=6(2t+5)
3−3+26=(2+5)6
3−3+26=(2+5)6
\frac{{\color{#c92786}{3t}}-3+{\color{#c92786}{2t}}}{6}=\frac{(2t+5)}{6}63t−3+2t=6(2t+5)
5−36=(2+5)6
Multiply all terms by the same value to eliminate fraction denominators
5−36=2+56
\frac{5t-3}{6}=\frac{2t+5}{6}65t−3=62t+5
6⋅5−36=6(2+56)
Cancel multiplied terms that are in the denominator
6⋅5−36=6(2+56)
6 \cdot \frac{5t-3}{6}=6(\frac{2t+5}{6})6⋅65t−3=6(62t+5)
5−3=2+5
5−3=2+5
5t-3=2t+55t−3=2t+5
5−3+3=2+5+3
5=2+8
5t=2t+85t=2t+8
5−2=2+8−2
3=8
3t=83t=8
33=8/3
=8/3