Solution 2.3:3e
From Förberedande kurs i matematik 1
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- | {{ | + | In this case, we see that the left-hand side contains the factor <math>x+3</math>, which we can take out to obtain |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | (x+3)(x-1) - (x+3)(2x-9) | ||
+ | &= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt] | ||
+ | &= (x+3)(x-1-2x+9)\\[5pt] | ||
+ | &= (x+3)(-x+8)\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | This rewriting of the equation results in the new equation | ||
+ | |||
+ | {{Displayed math||<math>(x+3)(-x+8)=0</math>}} | ||
+ | |||
+ | which has the solutions <math>x=-3</math> and <math>x=8\,</math>. | ||
+ | |||
+ | We check the solution <math>x=8</math> by substituting it into the equation, | ||
+ | |||
+ | {{Displayed math||<math>\text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.}</math>}} |
Current revision
In this case, we see that the left-hand side contains the factor \displaystyle x+3, which we can take out to obtain
\displaystyle \begin{align}
(x+3)(x-1) - (x+3)(2x-9) &= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt] &= (x+3)(x-1-2x+9)\\[5pt] &= (x+3)(-x+8)\,\textrm{.} \end{align} |
This rewriting of the equation results in the new equation
\displaystyle (x+3)(-x+8)=0 |
which has the solutions \displaystyle x=-3 and \displaystyle x=8\,.
We check the solution \displaystyle x=8 by substituting it into the equation,
\displaystyle \text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.} |