2x+5y-13=0 3x+5y=12

X=-1,y=3

Get the variable with the smaller coefficient by itself: 2x+5y−13=02x+5y−13=0 ⟹2x=−5y+13⟹2x=−5y+13 ⟹2x=−4y+12+1−y⟹2x=−4y+12+1−y ⟹2x2=−4y 2+122+1−y2⟹2x2=−4y 2+122+1−y2 ⟹x=−2y+6+1−y2.⟹x=−2y+6+1−y2. Next letting t=1−y2t=1−y2 ⟹2t=1−y⟹2t=1−y ⟹y=1−2t.⟹y=1−2t. Let tt be an integer, which means that yy is an integer. Then x=−2y+6+1−y2x=−2y+6+1−y2 =−2(1−2t)+6+t=−2(1−2t)+6+t =−2+4t+6+t=−2+4t+6+t =4+5t.=4+5t. Check these two answers: 2x+5y−13=02x+5y−13=0 ⟹2(4+5t)+5(1−2t)−13⟹2(4+5t)+5(1−2t)−13 =8+10t+5−10t−13=8+10t+5−10t−13 =13+10t−10t−13=0✓=13+10t−10t−13=0✓ ∴∴ integer solutions to the given equation have the form x=4+5t, y=1−2tx=4+5t, y=1−2t where tt is an integer. So there is a countable infinitude of integer solutions to the given equation and more than 4.4. Choose integer values of tt to obtain specific answers. For example, t=10t=10 yields the answer x=4+5(10)=4+50=54, y=1−2(10)=1−20=−19.x=4+5(10)=4+50=54, y=1−2(10)=1−20=−19. Select four different values of tt to get four distinct solutions. To obtain positive answers for xx and y,y, must have x=4+5t>0 and y=1−2t>0x=4+5t>0 and y=1−2t>0 ⟹5t>−4 and 1>2t⟹5t>−4 and 1>2t ⟹t>−45 and t<12⟹t>−45 and t<12 ⟹−45<t<12⟹−45<t<12 ⟹t=0⟹t=0 since tt is an integer. Then the only positive solution is x=4+5(0)=4, y=1−2(0)=1.