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2 years ago

How many solutions does the system have? 6x-y=-1 6x+y=-1

Mathematics

1 answer:

vitfil [10]2 years ago

5 0

This system only have one solution to the problem that is asked

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3x-9=3(x-3) how many solutions

Colt1911 [192]

Answer:

All Real Numbers

Step-by-step explanation:

Step 1:

3x - 9 = 3 ( x - 3 ) Equation

Step 2:

3x - 9 = 3x - 9 Multiply

Step 3:

- 9 = - 9 Subtract 3x on both sides

Answer:

All Real Numbers

Hope This Helps :)

4 0

2 years ago

Find the value of X if H is the midpoint of JG, GJ equals 4X -6 and GH equals 27.

kupik [55]

Answer:

Step-by-step explanation:

7 0

2 years ago

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

How i have to do it I don't understand

ZanzabumX [31]

I think the answer is 0.04...

4 0

2 years ago

If x = 1 is a common root of ax² +ax + 2 = 0 and x² + x + b = 0 , then ab =

vodka [1.7K]

Answer:

ab = 2

Step-by-step explanation:

Given equations

ax² +ax + 2 = 0

x² + x + b = 0

root of both the equation

x= 1

then we can plug in x = 1 in both the equation

ax² +ax + 2 = 0 x² + x + b = 0

a*1² +a*1 + 2 = 0 1² + 1 + b = 0

a +a + 2 = 0 1 + 1 + b = 0

2a + 2 = 0 2 + b = 0

2a = - 2 b = -2

a = -2/2 = -1

Thus,

a = -1

b = -2

a*b = -1*-2 = 2

ab = 2

6 0

2 years ago