(14, -8 ) and (7, -6)
2 answers:
You might be interested in
(-2,-5)
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.
Answer:
Step-by-step explanation:
Answer:
(i) (x +4)^3 = x^3 +12x^2 +48x +64
(ii) (2m +1)^3 = 8m^3 +12m^2 +6m +1
(iii) (a -2b)^3 = a^3 -6a^2b +12ab^2 -8b^3
Step-by-step explanation:
Powers of a binomial have a pattern. The coefficients of the terms of the expanded product are the numbers in the corresponding row of Pascal's Triangle. Here, you want the 3rd power, so the coefficients of interest are ...
1, 3, 3, 1
The pattern is ...
(a +b)^2 = a^2 +2ab +b^2
(a +b)^3 = a^3 +3a^2b +3ab^2 +b^3
The total of powers of the variable is equal to the exponent of the binomial. The powers of the first variable decrease to 0 while the powers of the second variable increase from zero.
When 'a' or 'b' is something with a coefficient, that whole term is raised to the appropriate power.
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i) (x+4)^3 = x^3 +3·4x^2 +3·4^2x +4^3
(x +4)^3 = x^3 +12x^2 +48x +64
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ii) (2m +1)^3 = (2m)^3 +3·1·(2m)^2 +3·1^2·(2m) +1^3
(2m +1)^3 = 8m^3 +12m^2 +6m +1
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iii) (a -2b)^3 = a^3 +3·a^2·(-2b) +3·a·(-2b)^2 +(-2b)^3
(a -2b)^3 = a^3 -6a^2b +12ab^2 -8b^3
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<em>Additional comment on Pascal's triangle</em>
The numbers in each row are the sums of the two numbers immediately above. For binomial expansion, you're only interested in the contents of a row. The "patterns" shown in the attachment are an extra added attraction, not particularly relevant to binomial expansion.
