Asked by Utibe
Solve the following system of equation:
2x+y=8
3x-y=5
Use both substitution and elimination method
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Answered by
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We solve
2x + y = 8
3x − y = 5
1) Substitution
From the first equation y = 8 − 2x. Substitute into the second:
3x − (8 − 2x) = 5
3x − 8 + 2x = 5
5x − 8 = 5
5x = 13
x = 13/5.
Then y = 8 − 2(13/5) = 8 − 26/5 = 40/5 − 26/5 = 14/5.
2) Elimination
Add the two equations to eliminate y:
(2x + y) + (3x − y) = 8 + 5
5x = 13
x = 13/5.
Substitute back into 2x + y = 8:
2(13/5) + y = 8 → 26/5 + y = 8 → y = 14/5.
Solution: (x, y) = (13/5, 14/5) (≈ (2.6, 2.8)). Check: 2(13/5)+14/5=26/5+14/5=40/5=8; 3(13/5)−14/5=39/5−14/5=25/5=5.
2x + y = 8
3x − y = 5
1) Substitution
From the first equation y = 8 − 2x. Substitute into the second:
3x − (8 − 2x) = 5
3x − 8 + 2x = 5
5x − 8 = 5
5x = 13
x = 13/5.
Then y = 8 − 2(13/5) = 8 − 26/5 = 40/5 − 26/5 = 14/5.
2) Elimination
Add the two equations to eliminate y:
(2x + y) + (3x − y) = 8 + 5
5x = 13
x = 13/5.
Substitute back into 2x + y = 8:
2(13/5) + y = 8 → 26/5 + y = 8 → y = 14/5.
Solution: (x, y) = (13/5, 14/5) (≈ (2.6, 2.8)). Check: 2(13/5)+14/5=26/5+14/5=40/5=8; 3(13/5)−14/5=39/5−14/5=25/5=5.
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