# Which is true of the infinite solutions of the inequalit x <0?The solutions will contain an infinite

###### Question:

The solutions will contain an infinite amount of integers,

The solutions will contain an intinite amount of negative numbers,

The solutions will contain an infinite amount of negative numbers and zero,

## Answers

The solutions will contain an infinite amount of negative numbers.

Step-by-step explanation:

Given x < 0

What does this mean?

First, carefully examine the sign in the inequality. It says less (which is different from less and equal)

Then explain it's meaning.

x < 0 just means that the solutions will contain an infinite amount of negative numbers but NOT zero. This defines the correct answer.

Step-by-step explanation:

< = more than

> = less than

So therefore,

D > 5 miles

Hope this helped !

2x + 6y = − 12 ;10x + 32y = −62

Now make sure to solve!

2x + 6y = −12 for x > 2x + 6y + −6y = −12 + −6y (Just add -6y to both sides)

2x = −6y −12

[tex]\frac{2x}{2} = \frac{-6y - 12}{2}[/tex]< Divide both sides by 2

x= − 3y − 6

Now we have to substitute > −3y − 6 for x in 10x + 32y = −62

10x + 32y = −62

10(−3y − 6) + 32y = −62

2y − 60 = −62 (We have to simplify both sides!)

2y − 60 + 60 = −62 + 60 (Now add 60 to both sides!)

2y = −2

[tex]\frac{2y}{2} = \frac{-2}{2}[/tex] (Make sure to divide both sides by 2)

We get y = -1 here.

Substitute −1 for y in x = −3y − 6

−3y − 6x

(−3)(−1) − 6

x=−3 (Simplify both sides)

We have 2 solutions > x = −3 and y = −1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I am not quite sure how to solve the second problem!

1) Option (3) is correct. The system has one solution.

2) Option (2) is correct. The solution is (4,0)

Step-by-step explanation:

1) Consider the given system of equation ,

2x +6y = -12 .....(1)

10x + 32y = -62 ....(2)

Here, [tex]a_1=2,b_1=6,c_1--12[/tex] and [tex]a_2=10,b_2=32,c_2=-62[/tex]

Find the ratio for [tex]\frac{a_1}{a_2},\frac{b_1}{b_2},\frac{c_1}{c_2}[/tex]

Thus, [tex]\frac{a_1}{a_2}=\frac{2}{10}=\frac{1}{5}[/tex]

[tex]\frac{b_1}{b_2}=\frac{6}{32}=\frac{3}{16}[/tex]

and [tex]\frac{c_1}{c_2}=\frac{-12}{-62}=\frac{3}{8}[/tex]

hence, it is clear from above ratios, that [tex]\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq\frac{c_1}{c_2}[/tex]

Thus, the system will have a unique solution.

We will solve the system ,

Multiply equation (1) by 5, we get,

10x + 30y = -60 ...(3)

Subtract equation (3) from (2),

10x + 32y -( 10x + 30y ) = -62-( -60 )

⇒ y = - 1

Put y = 1 in (1), we get,

2x +6(-1) = -12 ⇒ 2x= -12 +6 ⇒ x = -3.

Thus, (-3 , -1) is the solution

Hence the system has one solution.

2) Consider the given system of equation ,

2d + e = 8 ...........(1)

d - e = 4 ..........(2)

Add equation (1) and (2) , we get,

2d + e +(d -e) = 8+ 4

⇒ 2d +d = 12

⇒ 3d= 12

⇒ d = 4

Put d = 4 in (1) , we get

2d + e = 8 ⇒ 2(4)+e = 8 ⇒ e = 8 - 8 ⇒ e = 0.

Thus, The solution is (4,0).

1, 2, 5 and 6 are the answers.

Step-by-step explanation:

2 is correct because the arrow shows that the number line continues till infinity. 5 is correct because any fraction is possible as no restriction of integer value is placed. Any number less than 7 is not included. There are some confusions with 1 as it's not a solid color, So, I assume -7 is not included.

answer:

b

explanation:

Here... I only have one answer.

[tex]1. solve the system. {j+k=3j−k=7{j+k=3j−k=7 use the substitution method. the solution is (5, −2)(5,[/tex]

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

[tex]5x+4y+5z=-1[/tex]

[tex]x+y+2z=1[/tex]

[tex]2x+y-z=-3[/tex]

Firs we find determinant of system of equations

Let a matrix A=[tex]\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right][/tex] and B=[tex]\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right][/tex]

[tex]\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}[/tex]

[tex]\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0[/tex]

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply [tex]R_1\rightarrow R_1-4R_2[/tex] and [tex]R_3\rightarrow R_3-2R_2[/tex]

[tex]\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right][/tex]:[tex]\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right][/tex]

Apply[tex]R_2\rightarrow R_2-R_1[/tex]

[tex]\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right][/tex]:[tex]\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right][/tex]

Apply [tex]R_3\rightarrow R_3+R_2[/tex]

[tex]\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right][/tex]:[tex]\left[\begin{array}{ccc}-5\\6\\1\end{array}\right][/tex]

Apply [tex]R_3\rightarrow- \frac{1}{2}[/tex] and [tex]R_2\rightarrow R_2-R_3[/tex]

[tex]\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right][/tex]:[tex]\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right][/tex]

Apply [tex]R_1\rightarrow R_1-R_3[/tex]

[tex]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right][/tex]:[tex]\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right][/tex]

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

It's C) x + 1 < 5; x < 4

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