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