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3 years ago

Solve for x: 3 - (2x - 5) < -4(x + 2)

Mathematics

2 answers:

sattari [20]3 years ago

5 0

x < -23/2

Step-by-step explanation:

you open the bracket first

LHS

3-(2x - 5)

-6x+15

RHS

-4(x+2)

=-4x-8

therefore:

-6x+15 < -4x-8

By collecting the like terms

-6x+4x < -8-15

=-2x<-23

Divide both side by -2x

-2x/-2 < -23/-2

x < -23/2

nata0808 [166]3 years ago

3 0

Step-by-step explanation:

3-(2x-5) <-4(x+2)

=> 3- 2x +5 +4x +8<0

=> 16 + 2x <0

=> 2x < -16

=> x < -8

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The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

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

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

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 $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

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

Apply $R_2\rightarrow R_2-R_1$

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

Apply $R_3\rightarrow R_3+R_2$

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

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

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

Apply $R_1\rightarrow R_1-R_3$

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

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.

4 0

3 years ago

Find the midpoint of points A(8,-7) and B(3,-1) graphically

sukhopar [10]

Answer: (5.5, -4)

Step-by-step explanation:

You can use the midpoint formula to find the midpoint of point A and point B.

$M(\frac{x_{1}+x_{2} }{2}, \frac{y_{1} + y_{2} }{2})$

x₁ = 8

x₂ = 3

y₁ = -7

y₂ = -1

$M(\frac{x_{1}+x_{2} }{2}, \frac{y_{1} + y_{2} }{2})$

$M(\frac{8+3 }{2}, \frac{-7 +-1}{2} )$

$M(\frac{8+3 }{2}, \frac{-7 -1}{2} )$

$M(\frac{11 }{2}, \frac{-8}{2} )$

$M(5.5, -4 )$

Midpoint = (5.5, -4)

Hope this helps!

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Answer: ....

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