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dsp73
4 years ago
7

What is a number line for 5a + 18 < -27?

Mathematics
2 answers:
devlian [24]4 years ago
4 0

Answer:

Letter C.

Step-by-step explanation:

Let’s start by subtracting \blue{18}18start color #6495ed, 18, end color #6495ed from both sides of the inequality:

\qquad \begin{aligned} 5a + 18 &< -27\\ 5a + 18 \blue{-18} &< -27\blue{-18}\\ 5a &< -45\\ \end{aligned}

5a+18

5a+18−18

5a​  

<−27

<−27−18

<−45

Hint #22 / 4

To isolate aaa, we need to divide both sides by \green{5}5start color #28ae7b, 5, end color #28ae7b:

\qquad\begin{aligned} 5a &< -45\\ \\ \dfrac{5a}{\green{5}} &< \dfrac{-45}{\green{5}}\\ \\ a &< \purple{-9}\\ \end{aligned}

5a

5

5a

a

<−45

<

5

−45

<−9

​Hint #33 / 4

To graph the inequality a < \purple{-9}a<−9a, is less than, start color #9d38bd, minus, 9, end color #9d38bd, we first draw a circle at \purple{-9}−9start color #9d38bd, minus, 9, end color #9d38bd. This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

Since the solution uses a less than sign, the solution does not include the point where a= \purple{-9}a=−9a, equals, start color #9d38bd, minus, 9, end color #9d38bd. So the circle at \purple{-9}−9start color #9d38bd, minus, 9, end color #9d38bd is not filled in.

Because the solution to the inequality says that a < \purple{-9}a<−9a, is less than, start color #9d38bd, minus, 9, end color #9d38bd, this means that solutions are numbers to the left of \purple{-9}−9start color #9d38bd, minus, 9, end color #9d38bd.

Hint #44 / 4

The graph that represents the solution of the inequality a < \purple{-9}a<−9a, is less than, start color #9d38bd, minus, 9, end color #9d38bd is shown in

Alexandra [31]4 years ago
3 0

a < -9

Rearrange:

Rearrange the equation by subtracting what is to the right of the greater than sign from both sides of the inequality :

                    5*a+18-(-27)<0  

Step by step solution :

Step  1  :

Pulling out like terms :

1.1     Pull out like factors :

  5a + 45  =   5 • (a + 9)  

Equation at the end of step  1  :

Step  2  :

2.1    Divide both sides by  5  

Solve Basic Inequality :

2.2      Subtract  9  from both sides

           a < -9

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Answer:

<em>(a) x=2, y=-1</em>

<em>(b)  x=2, y=2</em>

<em>(c)</em> \displaystyle x=\frac{5}{2}, y=\frac{5}{4}

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.

We call the determinant of the system

\Delta=\begin{vmatrix}a &b \\c  &d \end{vmatrix}

We also define:

\Delta_x=\begin{vmatrix}p &b \\q  &d \end{vmatrix}

And

\Delta_y=\begin{vmatrix}a &p \\c  &q \end{vmatrix}

The solution for x and y is

\displaystyle x=\frac{\Delta_x}{\Delta}

\displaystyle y=\frac{\Delta_y}{\Delta}

(a) The system to solve is

\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.

Calculating:

\Delta=\begin{vmatrix}1 &1 \\1  &-2 \end{vmatrix}=-2-1=-3

\Delta_x=\begin{vmatrix}1 &1 \\4  &-2 \end{vmatrix}=-2-4=-6

\Delta_y=\begin{vmatrix}1 &1 \\1  &4 \end{vmatrix}=4-3=3

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1

The solution is x=2, y=-1

(b) The system to solve is

\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.

Calculating:

\Delta=\begin{vmatrix}4 &-1 \\1  &-1 \end{vmatrix}=-4+1=-3

\Delta_x=\begin{vmatrix}6 &-1 \\0  &-1 \end{vmatrix}=-6-0=-6

\Delta_y=\begin{vmatrix}4 &6 \\1  &0 \end{vmatrix}=0-6=-6

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2

The solution is x=2, y=2

(c) The system to solve is

\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.

Calculating:

\Delta=\begin{vmatrix}-1 &2 \\1  &2 \end{vmatrix}=-2-2=-4

\Delta_x=\begin{vmatrix}0 &2 \\5  &2 \end{vmatrix}=0-10=-10

\Delta_y=\begin{vmatrix}-1 &0 \\1  &5 \end{vmatrix}=-5-0=-5

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}

The solution is

\displaystyle x=\frac{5}{2}, y=\frac{5}{4}

(d) The system to solve is

\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.

Calculating:

\Delta=\begin{vmatrix}6 &-1 \\4  &-2 \end{vmatrix}=-12+4=-8

\Delta_x=\begin{vmatrix}-5 &-1 \\6  &-2 \end{vmatrix}=10+6=16

\Delta_y=\begin{vmatrix}6 &-5 \\4  &6 \end{vmatrix}=36+20=56

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7

The solution is x=-2, y=-7

4 0
3 years ago
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