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4 years ago

find all whole number solutions of each system using a table.. 9.) x+y<8. 3x less than or equal to y+6. . . PLEASE HELP

2 answers:

7 0

We aare given with two inqualities x + y <8 and 3x ≤ y + 6. To solve these inequalities, we replace the less than sign by equal sings. We just transpose y in the second equation to the left side and ad the two equations resulting to x equal to numbers less than 3.5 and y equal to numbers less than 4.5.

Alona [7]4 years ago

3 0

The correct answers are:

x ≤ 3 and y < 4.

Explanation:

We can isolate the variable y in the first inequality by subtracting x from each side:

x+y < 8

x+y-x < 8-x

y < 8-x

Now we can substitute this into the second inequality:

3x ≤ y + 6

3x ≤ 8-x+6

Combining like terms on the right, we have

3x ≤ 14-x

We will add x to each side so that we only have a variable on one side of the inequality:

3x+x ≤ 14-x+x

4x ≤ 14

Divide both sides by 4:

4x/4 ≤ 14/4

x ≤ 3.5

Now we substitute this into the first inequality:

3.5+y<8

Subtract 3.5 from each side:

3.5+y-3.5 < 8-3.5

y < 4.5

Our solutions are x ≤ 3.5 and y < 4.5. However, we are asked to find the whole number solutions; this involves some rounding. While 3.5 would round up to 4, it will not work in our inequality, since x is to be less than or equal to 3.5 and 4 is neither less than nor equal to this number. Thus we say that x is less than or equal to 3.

4.5 would round up to 5; however, it will not work either, since y is to be less than 4.5 and 5 is not less than 4.5. Thus we say that x is less than 4.

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3 years ago

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4 years ago

Show that if S1 and S2 are subsets of a vector space V such that S1 c S2 then span(S1) c span(S2). In particular, if S1 c S2 the

klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

If, additionally $S_2\subseteq S_1$ then reversing the roles of S1 and S2 in the previous proof, $span(S_2)\subseteq span(S_1)$ . Then $span(S_1)\subseteq span(S_2)\subseteq span(S_1)$ , therefore $span(S_1)=span(S_2)$ .

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3 years ago