No they won’t be.Consider the linear combination (1)(u – v) + (1) (v – w) + (-1)(u – w).This will add to 0. But the coefficients aren’t all 0.Therefore, those vectors aren’t linearly independent.
You can try an example of this with (1, 0, 0), (0, 1, 0), and (0, 0, 1), the usual basis vectors of R3.
That method relied on spotting the solution immediately.If you couldn’t see that, then there’s another approach to the problem.
We know that u, v, w are linearly independent vectors.So if au + bv + cw = 0, then a, b, and c are all 0 by definition.
Suppose we wanted to ask whether u – v, v – w, and u – w are linearly independent.Then we’d like to see if there are non-zero coefficients in the linear combinationd(u – v) + e(v – w) + f(u – w) = 0, where d, e, and f are scalars.
Distributing, we get du – dv + ev – ew + fu – fw = 0.Then regrouping by vector: (d + f)u + (-d +e)v + (-e – f)w = 0.
But now we have a linear combo of u, v, and w vectors.Therefore, all the coefficients must be 0.So d + f = 0, -d + e = 0, and –e – f = 0. It turns out that there’s a free variable in this solution.Say you let d be the free variable.Then we see f = -d and e = d.
Then any solution of the form (d, e, f) = (d, d, -d) will make (d + f)u + (-d +e)v + (-e – f)w = 0 a true statement.
Let d = 1 and you get our original solution. You can let d = 2, 3, or anything if you want.
Answer: The father is 37 years and the son is 7 years
Step-by-step explanation:
Let x represent the present age of the father.
Let y represent the present age of the son.
Three years hence a father will four times as old as his son will be. This means that
x + 3 = 4(y + 3)
x + 3 = 4y + 12
x - 4y = 12 - 3
x - 4y = 9- - - - - - - - - - - - - - - -1
Before two years he was seven times as old as his son was. This means that
(x - 2) = 7(y - 2)
x - 2 = 7y - 14
x - 7y = - 14 + 2
x - 7y = - 12- - - - - - - - - - - - - - -2
Subtracting equation 2 from equation 1, it becomes
3y = 21
y = 21.3 = 7
Substituting y = 7 into equation 1, it becomes
x - 4 × 7 = 9
x - 28 = 9
x = 9 + 28
x = 37
Simplify
1
4
(
4
+
x
)
4
1
(4+x) to
4
+
x
4
4
4+x
.
4
+
x
4
=
4
3
4
4+x
=
3
4
2 Simplify
4
+
x
4
4
4+x
to
1
+
x
4
1+
4
x
.
1
+
x
4
=
4
3
1+
4
x
=
3
4
3 Subtract
1
1 from both sides.
x
4
=
4
3
−
1
4
x
=
3
4
−1
4 Simplify
4
3
−
1
3
4
−1 to
1
3
3
1
.
x
4
=
1
3
4
x
=
3
1
5 Multiply both sides by
4
4.
x
=
1
3
×
4
x=
3
1
×4
6 Simplify
1
3
×
4
3
1
×4 to
4
3
3
4
.
x
=
4
3
x=
3
4
three and twenty-seven thousand, five hundred twenty-five hundred-thousandths
Sequence 1,5,9,13,...
A(0) = 1 +4x0=1
A(1) =1 + 4x1 = 5
A(2) = 1+ 4x2= 9
A(3) = 1 +4x3=13
A(n)= 1+4n A(n-1) = 1+4(n-1) =1+4n-4= - 3+4n
A(n) - A(n-1) = (1+4n) - (-3=4n) = 4
A(n) = A(n-1) +4; 29 is the answer
