Question

Solve the following equations by trial and error
method:

$\left(1\right)x-10=5$

$\left(2\right)\frac{a}{7}+8=6$

$\left(3\right)x^2+6=9$

$\left(4\right)19=x+13$

$\left(5\right)\frac{x}{4}=9$

$\left(6\right)2x+4=3x-4$

Answer 1

$\left(1\right)$

$x-10=5\Rightarrow x=$ should be multiple of $5$

$x=5$ $-5\ne 5$

$x=10$ $0\ne 5$

$x=15$ $5=5$

$x=15$

$\left(2\right)$

$\frac{9}{7}=2$ a is multiple of $7$

$x=14$ $2=2$ $a=14$

$\left(3\right)$

$x^2=3$ $X=\sqrt{3}$

$\left(4\right)$

$19=x+13\Rightarrow x=6$

$\left(5\right)$

$\frac{x}{4}=9$ $x$ is multiple of $4$

$x=36$ $\frac{36}{4}=9$ $9=9$

$\left(6\right)$

$x=8$