Solving Simultaneous Equation By Elimination Method

Solving Simultaneous Equation By Elimination

Method

3x+2y=6

5x - y= 8

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other

3x+2y=6, 5x−y=8

5×3x+5×2y=5×6,3×5x+3(−1)y=3×8

Simplify

15x+10y=30,15x−3y=24

Subtract 15x−3y=24 from $15 x + 10 y = 30$ by subtracting like terms on each side of the equal sign

15x−15x+10y+3y=30−24

Cancel out 15x - 15x to get

10y+3y=30−24

Add $10 y$ to $3 y$

13y=30−24

Subtract 24 from 30 to get

13y= 6

Divide both side by 13

y=136

5x−136=8

5x=13110

Divide through by 5

x=1322

x=1322,y=136

Comments

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Solving Simultaneous equation using matrix Method Question: Find the values of x and y in the equations below 3 x + 2 y = 6 5 x − y = 8 Using matrix method Put the equations in standard form and then use matrices to solve the system of equations. 3 x + 2 y = 6 , 5 x − y = 8 Write the equations in matrix form ( 3 5 2 − 1 ) ( x y ) = ( 6 8 ) Left multiply the equation by the inverse matrix of ( 3 5 2 − 1 ) . i n v e r s e ( ( 3 5 2 − 1 ) ) ( 3 5 2 − 1 ) ( x y ) = i n v e r s e ( ( 3 5 2 − 1 ) ) ( 6 8 ) The product of a matrix and its inverse is the identity matrix ( 1 0 0 1 ) ( x y ) = i n v e r s e ( ( 3 5 2 − 1 ) ) ( 6 8 ) Multiply the matrices on the left hand side of the equal sign. ( x y ) = i n v e r s e ( ( 3 5 2 − 1 ) ) ( 6 8 ) For the 2 × 2 matrix ( a c b d ) , the inverse matrix is ( a d − b c d a d − b c − c a d − b c − b a d − b c a ) , so the matrix equation can be ...

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How to calculate the common difference when the first term and the nth term is know. Question: The 28th term of an AP is -5. Find the common difference if its first term is 31. Solution T1 = a ( First term of the sequence) a = 31 T28 = -5 (28th term of the sequence) d = ? (common difference) Recall the nth term formula of Ap i.e Tn = a + (n-1)d Substitute 28 for n to get the following equation T28 = a + (28-1)d T28 = a + 27d Substitute T28 for -5 and 31 for a in the equation above -5 = 31 + 27d Subtract 31 from both sides of the equation -5 -31 = 27d -36 = 27d Divide through by 27 -36/27 = 27d/27 ...

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