Asked by raul
solve each system by substitution
1.
y=xt1
2x+y=7
2.
x=y-2
3x-y=6
3.
y=2xt3
5x-y=-3
1.
y=xt1
2x+y=7
2.
x=y-2
3x-y=6
3.
y=2xt3
5x-y=-3
Answers
Answered by
Henry
1. y = xti.
2x+y = 7.
Substitute xt1 for y:
2x+xt1 = 7
x(2+t1) = 7. Oops! we have 3 unknowns
but only 2 Eqs.
2. x = y-2.
3x-y = 6.
Substitute y-2 for x:
3(y-2)-y = 6
3y-6-y = 6
2y = 6+6 = 12
Y = 6.
X = y-2.
Substitute 6 for Y:
X = 6-2 = 4.
Solution set: (x,y) = (4,6).
2x+y = 7.
Substitute xt1 for y:
2x+xt1 = 7
x(2+t1) = 7. Oops! we have 3 unknowns
but only 2 Eqs.
2. x = y-2.
3x-y = 6.
Substitute y-2 for x:
3(y-2)-y = 6
3y-6-y = 6
2y = 6+6 = 12
Y = 6.
X = y-2.
Substitute 6 for Y:
X = 6-2 = 4.
Solution set: (x,y) = (4,6).
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