# Test: Linear Equations - Ambitious

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Question 1:   Solve: $\left(3x+4\right)\left(3x-2\right)+\left(4-3x\right)\left(4+3x\right)=\left(2x-1\right)\left(2x+1\right)-\left(4x-4\right)\left(x+2\right)$
$x=\frac{1}{5}$
$x=\frac{1}{4}$
$x=\frac{1}{10}$
$x=-\frac{1}{10}$
Question 2:   Solve: $\left(x-1\right)\left(x-2\right)-\left(5x-6\right)\left(2x-4\right)=5\left(x-8\right)-{\left(3x-4\right)}^{2}$
$x=2.7$
$x=3.2$
$x=2.8$
$x=2.5$
Question 3:   Solve: $\frac{3x+6}{4}-\frac{2x-6}{5}=\frac{x+7}{8}$
$x=-8\frac{1}{9}$
$x=-8$
$x=-7$
$x=-7\frac{8}{9}$
Question 4:   Solve: $\frac{x-4}{3}+\frac{6x+7}{2}=\frac{2x+5}{9}-\frac{3x+5}{6}$
$x=-\frac{44}{65}$
$x=-\frac{33}{67}$
$x=-\frac{57}{65}$
$x=-\frac{25}{67}$
Question 5:   Solve: $\left(3x+2\right)\left(x-5\right)-\left(5x-1\right)\left(2x-3\right)=5-\left(7x+4\right)\left(x-3\right)$
$x=-\frac{1}{3}$
$x=-2\frac{4}{13}$
$x=-\frac{8}{9}$
$x=-1\frac{5}{17}$
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