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zhenek [66]
3 years ago
10

Write a matrix to represent the system of equations.

Mathematics
2 answers:
ioda3 years ago
7 0

Answer:

7   -1   6  |  -8

7  -5   7   |  1

0    1    -4 | -5

Step-by-step explanation:

Note that there are four columns in this system, representing the variables x, y and z and constants.  So the given matrix pattern is appropriate:  It has three columns for the coefficients of x, y and z and one column for the constants.

We need only identify the coefficients and transfer them into the appropriate boxes of the matrix pattern.

We get:

7   -1   6   -8

7  -5   7     1

0    1    -4  -5

ozzi3 years ago
5 0

7 - 1 + 6 | 8

7 - 5 + 7 | 1

3 - 4 | -5

mark my answer as brainlist please

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3 years ago
3x+-2y=-6<br> 5x-2y=7<br> how do i solve this Elimination pls
tatuchka [14]

Answer:

x=\frac{13}{2}, y=\frac{51}{4}

Step-by-step explanation:

We\ are\ given,\\3x-2y=-6\\5x-2y=7\\Considering\ this\ system\ of\ equations,\\Let\ 3x-2y=-6\ be\ E_1\\Let\ 5x-2y=7\ be\ E_2,\\Hence,\\Multiplying\ E2\ with\ -1, we\ get:\\3x-2y=-6\\-1(5x-2y)=7*-1\\Hence,\\3x-2y=-6\\-5x+2y=-7\\Now,\\Lets\ add\ E_1\ and\ E_2\ together\\Hence,\\(3x-2y)+(-5x+2y)=(-6)+(-7)\\Hence,\\3x-2y-5x+2y=-13\\Arranging\ And\ Adding\ Like\ terms,\ we\ have:\\3x-5x-2y+2y=-13\\Hence,\\-2x+0y=-13\\-2x=-13\\x=\frac{-13}{-2}=\frac{13}{2}

Now,\ we\ can\ substitute\ x=\frac{13}{2}\ in\ E_1\ or\ E_2,\\But,\\Here,\ we'll\ substitute\ it\ in\ E1,\\Hence,\\Substituting\ x=\frac{13}{2}\ in\ E1,\ we\ have:\\3*\frac{13}{2}-2y=-6\\\frac{39}{2}-2y=-6\\-2y=-6- \frac{39}{2}\\-2y=\frac{-12-39}{2}\\-2y=\frac{-51}{2}\\y=\frac{-51}{2*-2}=\frac{51}{4}\\Hence,\\x=\frac{13}{2}, y=\frac{51}{4}

4 0
3 years ago
I keep getting different answers. Can somone explain how I would set it all up?
Nikitich [7]

Answer:

MY=80

Step-by-step explanation:

MX=MY ( since M is the midpoint)

2x+10=3x-5

10+5=3x-2x

15=x

MX+MY=MX

2x+10+3x-5=MY

2(15)+10+3(15)-5=MY

MY=80

6 0
3 years ago
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