0.2 Equations and inequalities: solving linear equations  (Page 2/2)

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Method: solving equations

The general steps to solve equations are:

1. Expand (Remove) all brackets.
2. "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.
3. Group all like terms together and simplify as much as possible.
4. Factorise if necessary.
5. Find the solution.
6. Substitute solution into original equation to check answer.

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

1. We are given $4-x=4$ and are required to solve for $x$ .

2. Since there are no brackets, we can start with grouping like terms and then simplifying.

3. $\begin{array}{ccc}\hfill 4-x& =& 4\hfill \\ \hfill -x& =& 4-4\phantom{\rule{1.em}{0ex}}\left(\mathrm{move}\mathrm{all}\mathrm{constant}\mathrm{terms}\left(\mathrm{numbers}\right)\mathrm{to}\mathrm{the}\mathrm{RHS}\left(\mathrm{right}\mathrm{hand}\mathrm{side}\right)\right)\hfill \\ \hfill -x& =& 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{group}\mathrm{like}\mathrm{terms}\mathrm{together}\right)\hfill \\ \hfill -x& =& 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill -x& =& 0\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}x& =& 0\hfill \end{array}$
4. Substitute solution into original equation:

$4-0=4$
$4=4$

Since both sides are equal, the answer is correct.

5. The solution of $4-x=4$ is $x=0$ .

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

1. We are given $4\left(2x-9\right)-4x=4-6x$ and are required to solve for $x$ .

2. We start with expanding the brackets, then grouping like terms and then simplifying.

3. $\begin{array}{ccc}\hfill 4\left(2x-9\right)-4x& =& 4-6x\hfill \\ \hfill 8x-36-4x& =& 4-6x\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{expand}\mathrm{the}\mathrm{brackets}\right)\hfill \\ \hfill 8x-4x+6x& =& 4+36\phantom{\rule{1.em}{0ex}}\mathrm{move}\mathrm{all}\mathrm{terms}\mathrm{with}\mathrm{x}\mathrm{to}\mathrm{the}\mathrm{LHS}\hfill \\ \hfill \mathrm{and}\mathrm{all}\mathrm{constant}\mathrm{terms}\mathrm{to}\mathrm{the}\mathrm{RHS}\mathrm{of}\mathrm{the}=\\ \hfill \left(8x-4x+6x\right)& =& \left(4+36\right)\phantom{\rule{1.em}{0ex}}\left(\mathrm{group}\mathrm{like}\mathrm{terms}\mathrm{together}\right)\hfill \\ \hfill 10x& =& 40\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill \frac{10}{10}x& =& \frac{40}{10}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{divide}\mathrm{both}\mathrm{sides}\mathrm{by}10\right)\hfill \\ \hfill x& =& 4\hfill \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill 4\left(2\left(4\right)-9\right)-4\left(4\right)& =& 4-6\left(4\right)\hfill \\ \hfill 4\left(8-9\right)-16& =& 4-24\hfill \\ \hfill 4\left(-1\right)-16& =& -20\hfill \\ \hfill -4-16& =& -20\hfill \\ \hfill -20& =& -20\hfill \end{array}$

Since both sides are equal to $-20$ , the answer is correct.

5. The solution of $4\left(2x-9\right)-4x=4-6x$ is $x=4$ .

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

1. We are given $\frac{2-x}{3x+1}=2$ and are required to solve for $x$ .

2. Since there is a denominator of ( $3x+1$ ), we can start by multiplying both sides of the equation by ( $3x+1$ ). But because division by 0 is not permissible, there is a restriction on a value for x. ( $x\ne \frac{-1}{3}$ )

3. $\begin{array}{ccc}\hfill \frac{2-x}{3x+1}& =& 2\hfill \\ \hfill \left(2-x\right)& =& 2\left(3x+1\right)\hfill \\ \hfill 2-x& =& 6x+2\phantom{\rule{1.em}{0ex}}\left(\mathrm{remove}/\mathrm{expand}\mathrm{brackets}\right)\hfill \\ \hfill -x-6x& =& 2-2\phantom{\rule{1.em}{0ex}}\mathrm{move}\mathrm{all}\mathrm{terms}\mathrm{containing}\mathrm{x}\mathrm{to}\mathrm{the}\mathrm{LHS}\hfill \\ \hfill \mathrm{and}\mathrm{all}\mathrm{constant}\mathrm{terms}\left(\mathrm{numbers}\right)\mathrm{to}\mathrm{the}\mathrm{RHS}.\\ \hfill -7x& =& 0\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill x& =& 0÷\left(-7\right)\hfill \\ \hfill therefore\phantom{\rule{1.em}{0ex}}x& =& 0\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathrm{zero}\mathrm{divided}\mathrm{by}\mathrm{any}\mathrm{number}\mathrm{is}0\hfill \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill \frac{2-\left(0\right)}{3\left(0\right)+1}& =& 2\hfill \\ \hfill \frac{2}{1}& =& 2\hfill \end{array}$

Since both sides are equal to 2, the answer is correct.

5. The solution of $\frac{2-x}{3x+1}=2$ is $x=0$ .

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

1. We are given $\frac{4}{3}x-6=7x+2$ and are required to solve for $x$ .

2. We start with multiplying each of the terms in the equation by 3, then grouping like terms and then simplifying.

3. $\begin{array}{ccc}\hfill \frac{4}{3}x-6& =& 7x+2\hfill \\ \hfill 4x-18& =& 21x+6\phantom{\rule{1.em}{0ex}}\left(\mathrm{each}\mathrm{term}\mathrm{is}\mathrm{multiplied}\mathrm{by}3\right)\hfill \\ \hfill 4x-21x& =& 6+18\phantom{\rule{1.em}{0ex}}\left(\mathrm{move}\mathrm{all}\mathrm{terms}\mathrm{with}\mathrm{x}\mathrm{to}\mathrm{the}\mathrm{LHS}\hfill \\ \hfill \mathrm{and}\mathrm{all}\mathrm{constant}\mathrm{terms}\mathrm{to}\mathrm{the}\mathrm{RHS}\mathrm{of}\mathrm{the}=\right)\\ \hfill -17x& =& 24\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill \frac{-17}{-17}x& =& \frac{24}{-17}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{divide}\mathrm{both}\mathrm{sides}\mathrm{by}-17\right)\hfill \\ \hfill x& =& \frac{-24}{17}\hfill \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill \frac{4}{3}×\frac{-24}{17}-6& =& 7×\frac{-24}{17}+2\hfill \\ \hfill \frac{4×\left(-8\right)}{\left(17\right)}-6& =& \frac{7×\left(-24\right)}{17}+2\hfill \\ \hfill \frac{\left(-32\right)}{17}-6& =& \frac{-168}{17}+2\hfill \\ \hfill \frac{-32-102}{17}& =& \frac{\left(-168\right)+34}{17}\hfill \\ \hfill \frac{-134}{17}& =& \frac{-134}{17}\hfill \end{array}$

Since both sides are equal to $\frac{-134}{17}$ , the answer is correct.

5. The solution of $\frac{4}{3}x-6=7x+2$ is,    $x=\frac{-24}{17}$ .

Solving linear equations

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

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