Solving linear equations (Page 2/2)

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Method: solving equations

The general steps to solve equations are:

Expand (Remove) all brackets.
"Move" all terms with the variable to the left hand side of the equation, and all constant terms (the numbers) to the right hand side of the equals sign.Bearing in mind that the sign of the terms will change from ( $+$ ) to ( $-$ ) or vice versa, as they "cross over" the equals sign.
Group all like terms together and simplify as much as possible.
Factorise if necessary.
Find the solution.
Substitute solution into original equation to check answer.

Khan academy video on equations - 1

Solve for $x$ : $4 - x = 4$

Determine what is given and what is required
We are given $4 - x = 4$ and are required to solve for $x$ .
Determine how to approach the problem
Since there are no brackets, we can start with grouping like terms and then simplifying.
Solve the problem
$\begin{matrix} 4 - x & = & 4 \\ - x & = & 4 - 4 (move all constant terms (numbers) to the RHS (right hand side)) \\ - x & = & 0 (group like terms together) \\ - x & = & 0 (simplify grouped terms) \\ - x & = & 0 \\ ∴ x & = & 0 \end{matrix}$
Check the answer
Substitute solution into original equation:

$4 - 0 = 4$

$4 = 4$

Since both sides are equal, the answer is correct.
Write the final answer
The solution of $4 - x = 4$ is $x = 0$ .

Solve for $x$ : $4 (2 x - 9) - 4 x = 4 - 6 x$

Determine what is given and what is required
We are given $4 (2 x - 9) - 4 x = 4 - 6 x$ and are required to solve for $x$ .
Determine how to approach the problem
We start with expanding the brackets, then grouping like terms and then simplifying.
Solve the problem
$\begin{matrix} 4 (2 x - 9) - 4 x & = & 4 - 6 x \\ 8 x - 36 - 4 x & = & 4 - 6 x (expand the brackets) \\ 8 x - 4 x + 6 x & = & 4 + 36 move all terms with x to the LHS \\ and all constant terms to the RHS of the = \\ (8 x - 4 x + 6 x) & = & (4 + 36) (group like terms together) \\ 10 x & = & 40 (simplify grouped terms) \\ \frac{10}{10} x & = & \frac{40}{10} (divide both sides by 10) \\ x & = & 4 \end{matrix}$
Check the answer
Substitute solution into original equation:

$\begin{matrix} 4 (2 (4) - 9) - 4 (4) & = & 4 - 6 (4) \\ 4 (8 - 9) - 16 & = & 4 - 24 \\ 4 (- 1) - 16 & = & - 20 \\ - 4 - 16 & = & - 20 \\ - 20 & = & - 20 \end{matrix}$

Since both sides are equal to $- 20$ , the answer is correct.
Write the final answer
The solution of $4 (2 x - 9) - 4 x = 4 - 6 x$ is $x = 4$ .

Solve for $x$ : $\frac{2 - x}{3 x + 1} = 2$

Determine what is given and what is required
We are given $\frac{2 - x}{3 x + 1} = 2$ and are required to solve for $x$ .
Determine how to approach the problem
Since there is a denominator of ( $3 x + 1$ ), we can start by multiplying both sides of the equation by ( $3 x + 1$ ). But because division by 0 is not permissible, there is a restriction on a value for x. ( $x \neq \frac{- 1}{3}$ )
Solve the problem
$\begin{matrix} \frac{2 - x}{3 x + 1} & = & 2 \\ (2 - x) & = & 2 (3 x + 1) \\ 2 - x & = & 6 x + 2 (remove / expand brackets) \\ - x - 6 x & = & 2 - 2 move all terms containing x to the LHS \\ and all constant terms (numbers) to the RHS . \\ - 7 x & = & 0 (simplify grouped terms) \\ x & = & 0 \div (- 7) \\ t h e r e f o r e x & = & 0 zero divided by any number is 0 \end{matrix}$
Check the answer
Substitute solution into original equation:

$\begin{matrix} \frac{2 - (0)}{3 (0) + 1} & = & 2 \\ \frac{2}{1} & = & 2 \end{matrix}$

Since both sides are equal to 2, the answer is correct.
Write the final answer
The solution of $\frac{2 - x}{3 x + 1} = 2$ is $x = 0$ .

Solve for $x$ : $\frac{4}{3} x - 6 = 7 x + 2$

Determine what is given and what is required
We are given $\frac{4}{3} x - 6 = 7 x + 2$ and are required to solve for $x$ .
Determine how to approach the problem
We start with multiplying each of the terms in the equation by 3, then grouping like terms and then simplifying.
Solve the problem
$\begin{matrix} \frac{4}{3} x - 6 & = & 7 x + 2 \\ 4 x - 18 & = & 21 x + 6 (each term is multiplied by 3) \\ 4 x - 21 x & = & 6 + 18 (move all terms with x to the LHS \\ and all constant terms to the RHS of the =) \\ - 17 x & = & 24 (simplify grouped terms) \\ \frac{- 17}{- 17} x & = & \frac{24}{- 17} (divide both sides by - 17) \\ x & = & \frac{- 24}{17} \end{matrix}$
Check the answer
Substitute solution into original equation:

$\begin{matrix} \frac{4}{3} \times \frac{- 24}{17} - 6 & = & 7 \times \frac{- 24}{17} + 2 \\ \frac{4 \times (- 8)}{(17)} - 6 & = & \frac{7 \times (- 24)}{17} + 2 \\ \frac{(- 32)}{17} - 6 & = & \frac{- 168}{17} + 2 \\ \frac{- 32 - 102}{17} & = & \frac{(- 168) + 34}{17} \\ \frac{- 134}{17} & = & \frac{- 134}{17} \end{matrix}$

Since both sides are equal to $\frac{- 134}{17}$ , the answer is correct.
Write the final answer
The solution of $\frac{4}{3} x - 6 = 7 x + 2$ is, $x = \frac{- 24}{17}$ .

Solving linear equations

Solve for $y$ : $2 y - 3 = 7$
Solve for $w$ : $- 3 w = 0$
Solve for $z$ : $4 z = 16$
Solve for $t$ : $12 t + 0 = 144$
Solve for $x$ : $7 + 5 x = 62$
Solve for $y$ : $55 = 5 y + \frac{3}{4}$
Solve for $z$ : $5 z = 3 z + 45$
Solve for $a$ : $23 a - 12 = 6 + 2 a$
Solve for $b$ : $12 - 6 b + 34 b = 2 b - 24 - 64$
Solve for $c$ : $6 c + 3 c = 4 - 5 (2 c - 3)$
Solve for $p$ : $18 - 2 p = p + 9$
Solve for $q$ : $\frac{4}{q} = \frac{16}{24}$
Solve for $q$ : $\frac{4}{1} = \frac{q}{2}$
Solve for $r$ : $- (- 16 - r) = 13 r - 1$
Solve for $d$ : $6 d - 2 + 2 d = - 2 + 4 d + 8$
Solve for $f$ : $3 f - 10 = 10$
Solve for $v$ : $3 v + 16 = 4 v - 10$
Solve for $k$ : $10 k + 5 + 0 = - 2 k + - 3 k + 80$
Solve for $j$ : $8 (j - 4) = 5 (j - 4)$
Solve for $m$ : $6 = 6 (m + 7) + 5 m$

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Source: OpenStax, Siyavula textbooks: grade 10 maths [ncs]. OpenStax CNX. Aug 05, 2011 Download for free at http://cnx.org/content/col11239/1.2

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