Mathematics, 31.12.2019 00:31 fgaef7234

Me. how do you form an augmented matrix from the system of linear equations and solve.

this is algebra 2. i need solving questions 6.3 & 6.4. i've read the instructions and i am not understanding what their asking me.

Me. how do you form an augmented matrix from the system of linear equations and solve.th

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Answer from: ehehvevsbg

6.3 (x, y) = (3, 1)

6.4 (x, y) = (-4, 2)

Step-by-step explanation:

An augmented matrix in this context is an array of the coefficients in the equation. For example, the equation x -2y = 3 can be represented by the row of numbers 1 -2 3. When writing the matrix by hand, we usually just separate the elements in a row using blank space. Some calculators require a comma or other character between matrix elements. The entire array is enclosed in square brackets.

An augmented matrix may have a vertical line separating the square matrix of coefficients from the rightmost column of constants.

Many graphing calculators have functions available for entering and solving linear equations in this form.

6.3 The augmented matrix is ...

$\left[\begin{array}{cc|c}2&3&9\\1&-4&-1\end{array}\right]$

This array can be used to solve the system of equations using "row operations." The idea is that any row can be multiplied by any number, and any row can be added to any other row (multiplied by some number or not). So, the same techniques used to solve a system by elimination can be used here.

We can, for example, subtract twice the second row from the first. Then we have ...

$\left[\begin{array}{cc|c}2-2(1)&3-2(-4)&9-2(-1)\\1&-4&-1\end{array}\right]\\\\\left[\begin{array}{cc|c}0&11&11\\1&-4&-1\end{array}\right] \quad\text{simplified}\\\\\left[\begin{array}{cc|c}0&1&1\\1&-4&-1\end{array}\right] \quad\text{first row divided by 11}\\\\\left[\begin{array}{cc|c}0&1&1\\1+4(0)&-4+4(1)&-1+4(1)\end{array}\right] \quad\text{4 times row 1 added to row 2}\\\\\left[\begin{array}{cc|c}0&1&1\\1&0&3\end{array}\right] \quad\text{simplified}$

When we turn this last matrix back into equations, we get ...

0x +y = 1

1x + 0y = 3

That is, the solution is (x, y) = (3, 1).

6.4 The augmented matrix is ...

$\left[\begin{array}{cc|c}4&5&-6\\3&2&-8\end{array}\right]$

Let's start the solution of this one by dividing the first row by 4.

$\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\3&2&-8\end{array}\right] \\\\\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\3-3(1)&2-3(\frac{5}{4})&-8-3(-\frac{3}{2})\end{array}\right] \quad\text{row 2 minus 3 times row 1}\\\\\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\0&-\frac{7}{4}&-\frac{7}{2}\end{array}\right] \quad\text{simplify}\\\\\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\0&1&2\end{array}\right] \quad\text{multiply row 2 by -4/7}$

We can eliminate the y-term in the first row by subtracting 5/4 times the second row. This will put the matrix in the form that shows us the solution directly.

$\left[\begin{array}{ccc}1-\frac{5}{4}(0)&\frac{5}{4}-\frac{5}{4}(1)&-\frac{3}{2}-\frac{5}{4}(2)\\0&1&2\end{array}\right] \quad\text{row 1 minus 5/4 times row 2}\\\\\left[\begin{array}{ccc}1&0&-4\\0&1&2\end{array}\right]$

This tells us ...

x = -4

y = 2

So the solution is (x, y) = (-4, 2).

Answer from: Quest

yes

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8. 4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

Answer from: Quest

i need more info. are you looked to see there mph? how much they deliver? how far they go?

step-by-step explanation:

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