3x−2y+z=−15,−x+y+2z=−10,x−y−4z=143x-2y+z=-15,-x+y+2z=-10,x-y-4z=14

3x−2y+z=−15,−x+y+2z=−10,x−y−4z=143x-2y+z=-15,-x+y+2z=-10,x-y-4z=14

  • DREAMS FROM MY FATHER

Why is this month with his father so important to Obama? Which memories of his father are most…

Looking back on his life, Barack Obama can see that the one month he spent with his birth father was of key importance. It marked their only time together. Dr. Obama passed away in 1982, when his…

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1+12,1+34,1+78,1+1516,1+31321+12,1+34,1+78,1+1516,1+3132 Write an expression for the apparent nth term…

T1=1+12T1=1+12 T2=1+34T2=1+34 T3=1+78T3=1+78 T4=1+1516T4=1+1516 T5=1+3132T5=1+3132 Every term has a ‘1’ in its expression. So the nth term will also have ‘1’ as the beginning value. When we consider the…

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1,3,322,336,3424,351201,3,322,336,3424,35120 Write an expression for the apparent nth term of the…

Let us first write the first two terms as fractions as well. 11,31,322,336,3424,35120,…11,31,322,336,3424,35120,… In the numerator we have powers of 3 (30=130=1 and 31=331=3). In the denominator we have…

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1,3,1,3,1,3,11,3,1,3,1,3,1 Write an expression for the apparent nth term of the sequence. (assume…

Let us start with the sequence −1,1,−1,1,−1,1,…-1,1,-1,1,-1,1,… The nnth term of this sequence is (−1)n(-1)n (for odd nn we get −1-1 while for even nnwe get 11) If we add 22 to each term of this sequence…

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  • MATH

1,−1,1,−1,1,−1,11,-1,1,-1,1,-1,1 Write an expression for the apparent nth term of the sequence….

The nn-th term is an=(−1)n−1an=(-1)n-1 Let us check: a1=(−1)0=1a1=(-1)0=1 a2=(−1)1=−1a2=(-1)1=-1 a3=(−1)2=1a3=(-1)2=1 a4=(−1)3=−1a4=(-1)3=-1 The solution is obvious if we know that (−1)n=1(-1)n=1 when nn is even and (−1)n=−1(-1)n=-1…

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1,12,16,124,11201,12,16,124,1120 Write an expression for the apparent nth term of the sequence….

Given 1, 1/2, 1/6, 1/24, 1/120 In this sequence the numerator of each term is 1. The terms in the denominator follow the n! (factorial) pattern. The expression for the nth them of the sequence is…

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  • MATH

1,14,19,116,1251,14,19,116,125 Write an expression for the apparent nth term of the sequence….

Numerator is always 1 while the denominator is a square number, thus we can write the sequence as follows 112,122,132,142,152,…112,122,132,142,152,… From this we see that the nnth terms is an=1n2an=1n2

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13,29,427,88113,29,427,881 Write an expression for the apparent nth term of the sequence….

Numerator contains powers of 2 starting with 20=120=1 while the denominator contains powers of 3 starting with 31=3.31=3. Therefore, the nnth term is an=2n−13nan=2n-13n

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21,33,45,57,6921,33,45,57,69 Write an expression for the apparent nth term of the sequence….

The numerators are an arithmetic sequence with the difference 1, the denominators — of the difference 2. The resulting formula is an=n+12n−1an=n+12n-1 .

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  • MATH

12,−14,18,−11612,-14,18,-116 Write an expression for the apparent nth term of the sequence….

The expression for the apparent term is an=(−1)n−12nan=(-1)n-12n  a1=(−1)1−121=12a1=(-1)1-121=12 a2=(−1)2−122=−14a2=(-1)2-122=-14 a3=(−1)3−123=18a3=(-1)3-123=18 a4=(−1)4−124=−116a4=(-1)4-124=-116

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−23,34,−45,56,−67-23,34,-45,56,-67 Write an expression for the apparent nth term of the sequence….

Numerator starts with 2 and is increased by one in each term that follows (n+1).(n+1).Denominator starts with 3 and is increased by one in each term that follows (n+2).(n+2). Signs alternate between + and…

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  • MATH

0,3,8,15,24,….0,3,8,15,24,…. Write an expression for the apparent nth term of the sequence….

Let us compare this sequence with the following one 1,4,9,16,25,…1,4,9,16,25,…This is the sequence of square numbers n2n2 and each term of our sequence is one smaller then the terms in the above sequence….

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  • MATH

3,7,11,15,19,…3,7,11,15,19,… Write an expression for the apparent nth term of the sequence….

It is obvious that each term is greater by four than the previous term which implies that this is arithmetic sequence. And since the first term is 3 we have an=3+4(n−1)=3+4n−4=−1+4nan=3+4(n-1)=3+4n-4=-1+4n The nnth…

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  • MATH

an=(−1)n+1n2+1an=(-1)n+1n2+1 Write the first five terms of the sequence. (assume that n…

an=(−1)n+1n2+1an=(-1)n+1n2+1 a1=(−1)1+112+1=12a1=(-1)1+112+1=12 a2=(−1)2+122+1=−15a2=(-1)2+122+1=-15 a3=(−1)3+132+1=110a3=(-1)3+132+1=110 a4=(−1)4+142+1=−117a4=(-1)4+142+1=-117 a5=(−1)5+152+1=126a5=(-1)5+152+1=126

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  • MATH

an=(−1)n(nn+1)an=(-1)n(nn+1) Write the first five terms of the sequence. (assume that n begins…

an=(−1)n(nn+1)an=(-1)n(nn+1) a1=(−1)1(11+1)=(−1)(12)=−12a1=(-1)1(11+1)=(-1)(12)=-12 Plug in n=2, to get the 2nd term a2=(−1)2(22+1)=(−1)(−1)(23)=23a2=(-1)2(22+1)=(-1)(-1)(23)=23Plug in n=3, to get the 3rd term…

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  • MATH

an=n(n2−6)an=n(n2-6) Write the first five terms of the sequence. (assume that n begins with 1.)

a1=1(12−6)=−5a1=1(12-6)=-5 a2=2(22−6)=2(−2)=−4a2=2(22-6)=2(-2)=-4 a3=3(32−6)=3⋅3=9a3=3(32-6)=3⋅3=9 a4=4(42−6)=4⋅10=40a4=4(42-6)=4⋅10=40 a5=5(52−6)=5⋅19=95a5=5(52-6)=5⋅19=95

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  • MATH

an=n(n−1)(n−2)an=n(n-1)(n-2) Write the first five terms of the sequence. (assume that n begins…

a1=1(1−1)(1−2)=1⋅0⋅(−1)=0a1=1(1-1)(1-2)=1⋅0⋅(-1)=0 a2=2(2−1)(2−2)=2⋅1⋅0=0a2=2(2-1)(2-2)=2⋅1⋅0=0 a3=3(3−1)(3−2)=3⋅2⋅1=6a3=3(3-1)(3-2)=3⋅2⋅1=6 a4=4(4−1)(4−2)=4⋅3⋅2=24a4=4(4-1)(4-2)=4⋅3⋅2=24 a5=5(5−1)(5−2)=5⋅4⋅3=60a5=5(5-1)(5-2)=5⋅4⋅3=60

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  • MATH

an=1+(−1)nan=1+(-1)n Write the first five terms of the sequence. (assume that n begins with 1.)

a1=1+(−1)1=1−1=0a1=1+(-1)1=1-1=0 a2=1+(−1)2=1+1=2a2=1+(-1)2=1+1=2 a3=1+(−1)3=1−1=0a3=1+(-1)3=1-1=0 a4=1+(−1)4=1+1=2a4=1+(-1)4=1+1=2 a5=1+(−1)5=1−1=0a5=1+(-1)5=1-1=0 This sequence alternates between 0 and 2.

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  • MATH

an=23an=23 Write the first five terms of the sequence. (assume that n begins with 1.)

This is a constant sequence, meaning that all the terms are the same. Therefore, the first five terms are the same. In this case they are equal to 23.23. a1=23a1=23 a2=23a2=23 a3=23a3=23 a4=23a4=23…

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  • MATH

an=1n32an=1n32 Write the first five terms of the sequence. (assume that n begins with 1.)

a1=1132=11=1a1=1132=11=1 a2=1232=18–√=122–√a2=1232=18=122 a3=1332=127−−√=133–√a3=1332=127=133 a4=1432=164−−√=144–√a4=1432=164=144 a5=1532=1125−−−√=155–√a5=1532=1125=155

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  • MATH

an=2n3nan=2n3n Write the first five terms of the sequence. (assume that n begins with 1.)

a1=2131=23a1=2131=23 a2=2232=49a2=2232=49 a3=2333=827a3=2333=827 a4=2434=1681a4=2434=1681 a5=2535=32243a5=2535=32243

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  • MATH

an=(−1)nn2an=(-1)nn2 Write the first five terms of the sequence. (assume that n begins with 1.)

The first five terms are -1, 1/4, -1/9, 1/16 and -1/25.

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  • MATH

an=1+(−1)nnan=1+(-1)nn Write the first five terms of the sequence. (assume that n begins…

(−1)n(-1)n is 1 for even n and -1 for odd n. So 1+(−1)n1+(-1)n is 2 or 0, respectively. So the first five terms are 0, 1/1, 0, 2/4 and 0.

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  • MATH

an=6n3n2−1an=6n3n2-1 Write the first five terms of the sequence. (assume that n begins…

These terms are fractions, their numerators and denominators are easy to compute. The first five terms are 3, 12/11, 18/26, 24/47 and 30/74.

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  • MATH

an=nn+2an=nn+2 Write the first five terms of the sequence. (assume that n begins with 1.)

an=nn+2an=nn+2 a1=11+2=13a1=11+2=13 a2=22+2=24=12a2=22+2=24=12 a3=33+2=35a3=33+2=35 a4=44+2=46=23a4=44+2=46=23 a5=55+2=57a5=55+2=57 So, the first five terms of the sequence are 13,12,35,23,5713,12,35,23,57

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  • MATH

an=(12)nan=(12)n Write the first five terms of the sequence. (assume that n begins with 1.)

an=(12)nan=(12)n a1=(12)1=12a1=(12)1=12 a2=(12)2=14a2=(12)2=14 a3=(12)3=18a3=(12)3=18 a4=(12)4=116a4=(12)4=116 a5=(12)5=132a5=(12)5=132 So, the first five terms of the sequence are 12,14,18,116,13212,14,18,116,132

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  • MATH

an=(−2)nan=(-2)n Write the first five terms of the sequence. (assume that n begins with 1.)

an=(−2)nan=(-2)n a1=(−2)1=−2a1=(-2)1=-2 a2=(−2)2=−2⋅−2=4a2=(-2)2=-2⋅-2=4 a3=(−2)3=−2⋅−2⋅−2=−8a3=(-2)3=-2⋅-2⋅-2=-8 a4=(−2)4=−2⋅−2⋅−2⋅−2=16a4=(-2)4=-2⋅-2⋅-2⋅-2=16 a5=(−2)5=−2⋅−2⋅−2⋅−2⋅−2=−32a5=(-2)5=-2⋅-2⋅-2⋅-2⋅-2=-32 So, the first five terms of the sequence are −2,4,−8,16,−32-2,4,-8,16,-32

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  • MATH

an=2−13nan=2-13n Write the first five terms of the sequence. (assume that n begins with 1.)

3n3n is 3, 9, 27, 81 and 243. Therefore the first five terms are 2-1/3 = 1 and 2/3, 2-1/9 = 1 and 8/9, 2-1/27 = 1 and 26/27, 2-1/81 = 1 and 80/81, 2-1/243 = 1 and 242/243.

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  • MATH

an=4n−7an=4n-7 Write the first five terms of the sequence. (assume that n begins with 1.)

an=4n−7an=4n-7 a1=4⋅1−7=−3a1=4⋅1-7=-3 a2=4⋅2−7=8−7=1a2=4⋅2-7=8-7=1 a3=4⋅3−7=12−7=5a3=4⋅3-7=12-7=5 a4=4⋅4−7=16−7=9a4=4⋅4-7=16-7=9 a5=4⋅5−7=20−7=13a5=4⋅5-7=20-7=13 So, the first five terms of the sequence are −3,1,5,9,13-3,1,5,9,13

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  • MATH

2x+2y−z=2,x−3y+z=−28,−x+y=14.2x+2y-z=2,x-3y+z=-28,-x+y=14. Use matricies to solve the system of…

2x+2y−z=22x+2y-z=2 x−3y+z=−28x-3y+z=-28 −x+y=14-x+y=14 The equations can be written as, ⎡⎣⎢⎢21−12−31−11028⎤⎦⎥⎥[22-11-3128-110] R2→→R1+R2+R3 ⎡⎣⎢⎢22−1201−10012⎤⎦⎥⎥[22-120012-110] R1 →→ R1-R2…

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  • MATH

−x+y−z=−14,2x−y+z=21,3x+2y+z=19-x+y-z=-14,2x-y+z=21,3x+2y+z=19 Use matricies to solve the system of…

The augmented matrix is ⎡⎣⎢−1231−12−111−142119⎤⎦⎥[-11-1-142-112132119] On applying R1−>R1+R2R1->R1+R2 we get (means changing 1st row as the sum of of first and second row)…

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  • MATH

2x−y+3z=24,2y−z=14,7x−5y=6.2x-y+3z=24,2y-z=14,7x-5y=6. Use matricies to solve the system of equations…

2x−y+3z=242x-y+3z=24 2y−z=142y-z=14 7x−5y=67x-5y=6 Write the equations as, ⎡⎣⎢⎢207−12−53−10⎤⎦⎥⎥[2-1302-17-50] Make the pivot in the first column by dividing the First row by 2,…

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  • MATH

x−3z=−2,3x+y−2z=5,2x+2y+z=4x-3z=-2,3x+y-2z=5,2x+2y+z=4 Use matricies to solve the system of…

x−3z=−2x-3z=-2 3x+y−2z=53x+y-2z=5 2x+2y+z=42x+2y+z=4 The equations can be written as ⎡⎣⎢⎢132012−3−212⎤⎦⎥⎥[10-3231-2221] R2 →→ (R1+R3)-R2 ⎡⎣⎢⎢102012−30123⎤⎦⎥⎥[10-320103221] R3 →→ (R3-2R2)…

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  • MATH

x+y+4z=5,2x+y−z=9x+y+4z=5,2x+y-z=9 Use matricies to solve the system of equations (if…

The augmented matrix is [12114−159][114521-19] On applying R1−>2R1−R2R1->2R1-R2 we get [10114951][11450191] On applying R1−>R1−R2R1->R1-R2 we get [1001−5941][10-540191] The corresponding…

1 educator answer

  • MATH

x+2y+z=8,3x+7y+6z=26x+2y+z=8,3x+7y+6z=26 Use matricies to solve the system of equations (if…

The augmented matrix is [132716826][121837626] On applying R2−>R2−3R1R2->R2-3R1 we get [10211382][12180132] On applying R1−>R1−2R2R1->R1-2R2 we get [1001−5342][10-540132] The corresponding…

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  • MATH

x−3y=5,−2x+6y=−10x-3y=5,-2x+6y=-10 Use matricies to solve the system of equations (if possible)….

Given system of equations are , x – 3y = 5, -2x + 6y = -10 so the matrix A,B are as follows, A = 1 -3 -2 6 and B = 5 -10 so the augmented matrix is [AB] = 1 -3 5…

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  • MATH

8x−4y=7,5x+2y=18x-4y=7,5x+2y=1 Use matricies to solve the system of equations (if possible). Use…

Given system of equations are 8x−4y=7,5x+2y=18x-4y=7,5x+2y=1 so the matrices A and B are given as follows A = [85−42][8-452] B = [71][71] so the augmented matrix is [AB] =

1 educator answer

  • MATH

5x−5y=−5,−2x−3y=75x-5y=-5,-2x-3y=7 Use matricies to solve the system of equations (if possible)….

Given system of equations are 5x – 5y = -5, -2x – 3y = 7 so ,we get the matrices as A = [5−2−5−3][5-5-2-3] and B = [−57][-57] the augmented matrix [AB] = [5−2−5−3−57][5-5-5-2-37] on…

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  • MATH

−2x+6y=−22,x+2y=−9-2x+6y=-22,x+2y=-9 Use matricies to solve the system of equations (if possible)….

Given , −2x+6y=−22,x+2y=−9-2x+6y=-22,x+2y=-9 A = [−2162][-2612] and B = [−22−9][-22-9] so the augmented matrix is [A B]= [−2162−22−9][-26-2212-9] step 1. Divide row 1 by -2

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  • MATH

x−4y+3z−2w=9,3x−2y+z−4w=−13,x-4y+3z-2w=9,3x-2y+z-4w=-13, 

x−4y+3z−2w=9x-4y+3z-2w=9 3x−2y+z−4w=−133x-2y+z-4w=-13 −4x+3y−2z+w=−4-4x+3y-2z+w=-4 −2x+y−4z+3w=−10-2x+y-4z+3w=-10 The above system of equations can be represented by the coefficient matrix A and right hand side matrix b as follows:…

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  • MATH

3x+2y−z+w=0,x−y+4z+2w=25,3x+2y-z+w=0,x-y+4z+2w=25, −2x+y+2z−w=2,x+y+z+w=6-2x+y+2z-w=2,x+y+z+w=6 Use…

3x+2y−z+w=03x+2y-z+w=0 x−y+4z+2w=25x-y+4z+2w=25 −2x+y+2z−w=2-2x+y+2z-w=2 x+y+z+w=6x+y+z+w=6 The above system of equations can be represented by the coefficient matrix A and right hand side vector b as follows:…

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  • MATH

x+2y=0,x+y=6,3x−2y=8x+2y=0,x+y=6,3x-2y=8 Use matricies to solve the system of equations. Use…

The augmented matrix ⎡⎣⎢11321−2068⎤⎦⎥[1201163-28] On applying R2−>R1−R2R2->R1-R2we get ⎡⎣⎢10321−20−68⎤⎦⎥[12001-63-28] On applying R3−>3R1−R3R3->3R1-R3 we get ⎡⎣⎢1002180−6−8⎤⎦⎥[12001-608-8] On…

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  • MATH

−x+y=−22,3x+4y=4,4x−8y=32-x+y=-22,3x+4y=4,4x-8y=32 Use matricies to solve the system of equations….

The augmented matrix is ⎡⎣⎢−13414−8−22432⎤⎦⎥[-11-223444-832] On applying R1−>−R1R1->-R1 and R3−>R34R3->R34 we get (means inverse the sign of first row and divide the third row by 4)…

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  • MATH

3x−2y+z=−15,−x+y+2z=−10,x−y−4z=143x-2y+z=-15,-x+y+2z=-10,x-y-4z=14Use matricies to solve the system of…

3x−2y+z=−153x-2y+z=-15 −x+y+2z=−10-x+y+2z=-10 x−y−4z=14x-y-4z=14 A=⎡⎣⎢3−11−21−112−4⎤⎦⎥A=[3-21-1121-1-4] b=⎡⎣⎢−15−1014⎤⎦⎥b=[-15-1014] [A∣b]=⎡⎣⎢3−11−21−112−4−15−1014⎤⎦⎥[A∣b]=[3-21-15-112-101-1-414] Multiply 2nd Row by 3 and add Row 1…

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  • MATH

x+2y−3z=−28,4y+2z=0,−x+y−z=−5x+2y-3z=-28,4y+2z=0,-x+y-z=-5 Use matricies to solve the system of…

x+2y−3z=−28x+2y-3z=-28 4y+2z=04y+2z=0 −x+y−z=−5-x+y-z=-5 The equations in the matrix form can be written as, ⎡⎣⎢10−1241−32−1−280−5⎤⎦⎥[12-3-280420-11-1-5]Add Row 1 and Row 3 ⎡⎣⎢100243−32−4−280−33⎤⎦⎥[12-3-28042003-4-33] Multiply…

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  • MATH

−x+y=4,2x−4y=−34-x+y=4,2x-4y=-34 Use matricies to solve the system of equations. Use Gaussian…

−x+y=4-x+y=4 2x−4y=−342x-4y=-34 Write the system of equations as an augmented matrix. [−121∣4−4∣−34][-11∣42-4∣-34] Multiply the first line by a 2. [−222∣8−4∣−34][-22∣82-4∣-34] Eliminate the first column….

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  • MATH

3x−2y=−27,x+3y=133x-2y=-27,x+3y=13 Use matricies to solve the system of equations. Use Gaussian…

rsWe began by stating our equations as follows: 3x – 2y = -27 x + 3y = 13 (NB: Note that I wrote the values of the x variable below each other, values with the y variable below each other and the…

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  • MATH

2x+6y=16,2x+3y=72x+6y=16,2x+3y=7 Use matricies to solve the system of equations. Use Gaussian…

2x+6y=162x+6y=16 2x−3y=72x-3y=7 Write the system of equations as an augmented matrix. [226∣16−3∣7][26∣162-3∣7] Multiply the second line by -1. [2−26∣163∣−7][26∣16-23∣-7] Eliminate the first column….

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  • MATH

x+2y=7,2x+y=8x+2y=7,2x+y=8 Use matricies to solve the system of equations. Use Gaussian…

x+2y=7x+2y=7 2x+y=82x+y=8 [122∣71∣8][12∣721∣8] Multiply the first line by -2. [−22−4∣−141∣8][-2-4∣-1421∣8] Eliminate the first column. [−20−4∣−14−3∣−6][-2-4∣-140-3∣-6]−3y=−6-3y=-6 y=2y=2 −2x−4y=−14-2x-4y=-14 −2x−4(2)=−14-2x-4(2)=-14…

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  • MATH

⎡⎣⎢4−113−508−160:−12:25:−29⎤⎦⎥[4-5-1:-12-1106:25380:-29] Write the system of linear…

The system of equations represented by the augmented matrix is 4x-5y-1z=-12 -11x+6z=25 3x+8y=-29

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