Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
x + 5y = 4
2x – 5y = 2
2007-09-18 07:41:48 · 8 answers · asked by spanishfly_100 1 in Science & Mathematics ➔ Mathematics
3x = 6
x = 2
2 + 5y = 4
5y = 2
y = 2 / 5
x = 2 , y = 2 / 5 (unique)
2007-09-18 22:03:19 · answer #1 · answered by Como 7 · 2⤊ 0⤋
x + 5y = 4
2x – 5y = 2 Add these 2 equations together
3x + 0 = 6
x = 2 Substitute x into 1st equation
2 + 5y = 4
y = 2/5 Check answer using 2nd equation
2(2) – 5(2/5) = 2
4 – 2 = 2
2 = 2
(x, y) = (2, 2/5) QED
2007-09-18 08:03:36 · answer #2 · answered by Richard A 2 · 0⤊ 1⤋
1) x + 5y = 4
2) 2x – 5y = 2
Add the two equations to eliminate y
x + 2x + 5y - 5y = 4 + 2
3x = 6
x = 2
Use eq 1 to solve for y
2 + 5y = 4
5y = 2
y = 2/5
Use eq 2 to check
2(2) – 5(2/5) = 2
4 - 2 = 2
2 = 2
2007-09-18 07:51:48 · answer #3 · answered by kindricko 7 · 0⤊ 0⤋
add the equations
3x=6
x=2
sub back in
2+5y=4
5y=2
y=2/5
verify
2(2)-5(2/5)=4-2=2 ta da
2007-09-18 07:47:30 · answer #4 · answered by Kenneth H 3 · 0⤊ 0⤋
x + 5y = 4
2x – 5y = 2
----------------
3x = 6
x = 2
2 + 5y = 4
5y = 2
y = 2/5
Sol: (2, 2/5)
2007-09-18 07:46:02 · answer #5 · answered by Becky M 4 · 0⤊ 1⤋
(2,2/5)
2007-09-18 08:00:53 · answer #6 · answered by mike m 1 · 0⤊ 0⤋
(2,2/5)
2007-09-18 07:47:32 · answer #7 · answered by JdR 1 · 0⤊ 0⤋
(2,2/5)
2007-09-18 07:46:26 · answer #8 · answered by xandyone 5 · 0⤊ 0⤋