You know that a*5x+b*2x has to equal 4x and a*1+b*(-3)=11. If we assume a and b are constants, then our two equations are 5a+2b=4 and a-3b=11. Solving this system of equations, we get a=2 and b=-3
Daniel S.
tutor
a and b are constants, so (a*5) and (b*2) are the coefficients of the x terms. For example, if a=1 and b=2, then the equation would look like 5x+4x=4x, which cannot be true. So we need a*5 and b*2 to add to 4
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06/04/15
Joseph J.
Oh yeah thanks!
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06/04/15
Michael J.
What Daniel did was first distribute to get rid of parentheses. He obtained
5ax + a + 2bx - 3b = 4x + 11
Then he equated coefficients by their term types. On the left side and right side of equation, there are x terms and constant terms. Lets equate both of these terms.
x ------> 5a + 2b = 4 eq1
constant ------> a - 3b = 11 eq2
We now have a system of equations.
Multiply eq2 by 5. Keep eq1 the same.
5a + 2b = 4 eq1
5a - 15b = 55 eq2
We can subtract eq2 from eq1 to eliminate the a terms. This will allow us to solve for b.
17b = -51
b = -3
This is the value of b that Daniel obtained.
Now, substitute this value of b we just obtained into eq1, to solve for a.
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06/04/15
Joseph J.
Thanks again
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06/04/15
Daniel S.
tutor
You got it
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06/04/15
Joseph J.
06/04/15