Basic Algebra/Introduction to Basic Algebra Ideas/Chapter Test

< Basic Algebra < Introduction to Basic Algebra Ideas

Please post your test problems this way:

Problem:(answer,level of difficulty(easy, medium, hard))
Example 2x=6: (x=3,easy)

(5 + x)^2 where x=4 (easy)
3*3+4 (easy)
2x–(2x+3) where x=.5 (medium)

Section A (5 marks)

Evaluate the following expressions.

$24\div 6-7=$
$(5+3)\times 6=$
${\frac {16+14}{5}}-3=$
${\frac {36}{4-(-2)}}+5=$
$({\frac {2\times 27}{9}})^{2}=$

Section B (5 marks)

Evaluate the following expressions.

$9+3^{2}=$
$3+{\frac {20}{2^{2}}}=$
${\frac {2(3+4)}{)}}{94^{0}}=$
$20-2({\frac {96}{4^{2}}})=$
$-4+5({\frac {56+7}{3^{2}}})-1=$

Section C (5 marks)

Evaluate the following expressions when $w=4$ , $x=5$ , $y=6$ and $z=7$ .

$3x+2y+z=$
${\frac {w^{2}}{y+2}}+2z=$
$2({\frac {x^{2}+3}{w}})-(2y-z)=$
$(6-3)({\frac {2w-x}{z-4}})+(3y-4)=$
$3^{2}(2w^{0})({\frac {wx}{y-2}})-z=$

Section D (5 marks)

Evaluate the value of $x$ in the following equations with the given value of $y$ .

$4x+3y=18$ (given $y=2$ )
${\frac {2x}{y+1}}+4=20$ (given $y=3$ )
$2({\frac {2x-2}{y-3}}+8=6^{2}$ (given $y=7$ )
$3^{3}+2^{0}+3x-2y({\frac {18}{2}})=97$ (given $y=3$ )
$[5(2x-3)]^{2}=y+6$ (given $y=19$ )

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