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Alles samen. Gebruik stappenplan en ZRM!

  1. \(4(5x+\frac{2}{3})=-9x+\frac{2}{9}\)
  2. \(-2(-3x-\frac{5}{11})=-7x+\frac{2}{9}\)
  3. \(5(4x+\frac{3}{11})=-3x+\frac{4}{7}\)
  4. \(-2(-3x-\frac{5}{7})=5x+\frac{3}{5}\)
  5. \(2(-5x-\frac{3}{5})=-7x+\frac{7}{11}\)
  6. \(-5(4x-\frac{3}{4})=7x+\frac{6}{7}\)
  7. \(3(4x-\frac{5}{4})=-5x+\frac{7}{5}\)
  8. \(-2(3x+\frac{5}{7})=-7x+\frac{5}{8}\)
  9. \(4(-2x+\frac{2}{5})=-9x+\frac{2}{3}\)
  10. \(-5(3x+\frac{4}{3})=-4x+\frac{5}{2}\)
  11. \(-7(5x-\frac{3}{11})=-9x+\frac{2}{5}\)
  12. \(5(4x-\frac{4}{3})=-7x+\frac{6}{5}\)

Alles samen. Gebruik stappenplan en ZRM!

Verbetersleutel

  1. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (5x+\frac{2}{3})& = & -9x+\frac{2}{9} \\\Leftrightarrow & 20x+\frac{8}{3}& = & -9x+\frac{2}{9} \\ & & & \text{kgv van noemers 3 en 9 is 9} \\\Leftrightarrow & \color{blue}{9} .(\frac{180}{ \color{blue}{9} }x+ \frac{24}{ \color{blue}{9} })& = & (\frac{-81}{ \color{blue}{9} }x+ \frac{2}{ \color{blue}{9} }). \color{blue}{9} \\\Leftrightarrow & 180x \color{red}{+24} & = & \color{red}{-81x} +2 \\\Leftrightarrow & 180x \color{red}{+24} \color{blue}{-24} \color{blue}{+81x} & = & \color{red}{-81x} +2 \color{blue}{+81x} \color{blue}{-24} \\\Leftrightarrow & 180x+81x& = & 2-24 \\\Leftrightarrow & \color{red}{261} x& = & -22 \\\Leftrightarrow & x = \frac{-22}{261} & & \\ & V = \left\{ \frac{-22}{261} \right\} & \\\end{align}\)
  2. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (-3x-\frac{5}{11})& = & -7x+\frac{2}{9} \\\Leftrightarrow & 6x+\frac{10}{11}& = & -7x+\frac{2}{9} \\ & & & \text{kgv van noemers 11 en 9 is 99} \\\Leftrightarrow & \color{blue}{99} .(\frac{594}{ \color{blue}{99} }x+ \frac{90}{ \color{blue}{99} })& = & (\frac{-693}{ \color{blue}{99} }x+ \frac{22}{ \color{blue}{99} }). \color{blue}{99} \\\Leftrightarrow & 594x \color{red}{+90} & = & \color{red}{-693x} +22 \\\Leftrightarrow & 594x \color{red}{+90} \color{blue}{-90} \color{blue}{+693x} & = & \color{red}{-693x} +22 \color{blue}{+693x} \color{blue}{-90} \\\Leftrightarrow & 594x+693x& = & 22-90 \\\Leftrightarrow & \color{red}{1287} x& = & -68 \\\Leftrightarrow & x = \frac{-68}{1287} & & \\ & V = \left\{ \frac{-68}{1287} \right\} & \\\end{align}\)
  3. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (4x+\frac{3}{11})& = & -3x+\frac{4}{7} \\\Leftrightarrow & 20x+\frac{15}{11}& = & -3x+\frac{4}{7} \\ & & & \text{kgv van noemers 11 en 7 is 77} \\\Leftrightarrow & \color{blue}{77} .(\frac{1540}{ \color{blue}{77} }x+ \frac{105}{ \color{blue}{77} })& = & (\frac{-231}{ \color{blue}{77} }x+ \frac{44}{ \color{blue}{77} }). \color{blue}{77} \\\Leftrightarrow & 1540x \color{red}{+105} & = & \color{red}{-231x} +44 \\\Leftrightarrow & 1540x \color{red}{+105} \color{blue}{-105} \color{blue}{+231x} & = & \color{red}{-231x} +44 \color{blue}{+231x} \color{blue}{-105} \\\Leftrightarrow & 1540x+231x& = & 44-105 \\\Leftrightarrow & \color{red}{1771} x& = & -61 \\\Leftrightarrow & x = \frac{-61}{1771} & & \\ & V = \left\{ \frac{-61}{1771} \right\} & \\\end{align}\)
  4. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (-3x-\frac{5}{7})& = & 5x+\frac{3}{5} \\\Leftrightarrow & 6x+\frac{10}{7}& = & 5x+\frac{3}{5} \\ & & & \text{kgv van noemers 7 en 5 is 35} \\\Leftrightarrow & \color{blue}{35} .(\frac{210}{ \color{blue}{35} }x+ \frac{50}{ \color{blue}{35} })& = & (\frac{175}{ \color{blue}{35} }x+ \frac{21}{ \color{blue}{35} }). \color{blue}{35} \\\Leftrightarrow & 210x \color{red}{+50} & = & \color{red}{175x} +21 \\\Leftrightarrow & 210x \color{red}{+50} \color{blue}{-50} \color{blue}{-175x} & = & \color{red}{175x} +21 \color{blue}{-175x} \color{blue}{-50} \\\Leftrightarrow & 210x-175x& = & 21-50 \\\Leftrightarrow & \color{red}{35} x& = & -29 \\\Leftrightarrow & x = \frac{-29}{35} & & \\ & V = \left\{ \frac{-29}{35} \right\} & \\\end{align}\)
  5. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{2} (-5x-\frac{3}{5})& = & -7x+\frac{7}{11} \\\Leftrightarrow & -10x-\frac{6}{5}& = & -7x+\frac{7}{11} \\ & & & \text{kgv van noemers 5 en 11 is 55} \\\Leftrightarrow & \color{blue}{55} .(\frac{-550}{ \color{blue}{55} }x- \frac{66}{ \color{blue}{55} })& = & (\frac{-385}{ \color{blue}{55} }x+ \frac{35}{ \color{blue}{55} }). \color{blue}{55} \\\Leftrightarrow & -550x \color{red}{-66} & = & \color{red}{-385x} +35 \\\Leftrightarrow & -550x \color{red}{-66} \color{blue}{+66} \color{blue}{+385x} & = & \color{red}{-385x} +35 \color{blue}{+385x} \color{blue}{+66} \\\Leftrightarrow & -550x+385x& = & 35+66 \\\Leftrightarrow & \color{red}{-165} x& = & 101 \\\Leftrightarrow & x = \frac{101}{-165} & & \\\Leftrightarrow & x = \frac{-101}{165} & & \\ & V = \left\{ \frac{-101}{165} \right\} & \\\end{align}\)
  6. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-5} (4x-\frac{3}{4})& = & 7x+\frac{6}{7} \\\Leftrightarrow & -20x+\frac{15}{4}& = & 7x+\frac{6}{7} \\ & & & \text{kgv van noemers 4 en 7 is 28} \\\Leftrightarrow & \color{blue}{28} .(\frac{-560}{ \color{blue}{28} }x+ \frac{105}{ \color{blue}{28} })& = & (\frac{196}{ \color{blue}{28} }x+ \frac{24}{ \color{blue}{28} }). \color{blue}{28} \\\Leftrightarrow & -560x \color{red}{+105} & = & \color{red}{196x} +24 \\\Leftrightarrow & -560x \color{red}{+105} \color{blue}{-105} \color{blue}{-196x} & = & \color{red}{196x} +24 \color{blue}{-196x} \color{blue}{-105} \\\Leftrightarrow & -560x-196x& = & 24-105 \\\Leftrightarrow & \color{red}{-756} x& = & -81 \\\Leftrightarrow & x = \frac{-81}{-756} & & \\\Leftrightarrow & x = \frac{3}{28} & & \\ & V = \left\{ \frac{3}{28} \right\} & \\\end{align}\)
  7. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (4x-\frac{5}{4})& = & -5x+\frac{7}{5} \\\Leftrightarrow & 12x-\frac{15}{4}& = & -5x+\frac{7}{5} \\ & & & \text{kgv van noemers 4 en 5 is 20} \\\Leftrightarrow & \color{blue}{20} .(\frac{240}{ \color{blue}{20} }x- \frac{75}{ \color{blue}{20} })& = & (\frac{-100}{ \color{blue}{20} }x+ \frac{28}{ \color{blue}{20} }). \color{blue}{20} \\\Leftrightarrow & 240x \color{red}{-75} & = & \color{red}{-100x} +28 \\\Leftrightarrow & 240x \color{red}{-75} \color{blue}{+75} \color{blue}{+100x} & = & \color{red}{-100x} +28 \color{blue}{+100x} \color{blue}{+75} \\\Leftrightarrow & 240x+100x& = & 28+75 \\\Leftrightarrow & \color{red}{340} x& = & 103 \\\Leftrightarrow & x = \frac{103}{340} & & \\ & V = \left\{ \frac{103}{340} \right\} & \\\end{align}\)
  8. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (3x+\frac{5}{7})& = & -7x+\frac{5}{8} \\\Leftrightarrow & -6x-\frac{10}{7}& = & -7x+\frac{5}{8} \\ & & & \text{kgv van noemers 7 en 8 is 56} \\\Leftrightarrow & \color{blue}{56} .(\frac{-336}{ \color{blue}{56} }x- \frac{80}{ \color{blue}{56} })& = & (\frac{-392}{ \color{blue}{56} }x+ \frac{35}{ \color{blue}{56} }). \color{blue}{56} \\\Leftrightarrow & -336x \color{red}{-80} & = & \color{red}{-392x} +35 \\\Leftrightarrow & -336x \color{red}{-80} \color{blue}{+80} \color{blue}{+392x} & = & \color{red}{-392x} +35 \color{blue}{+392x} \color{blue}{+80} \\\Leftrightarrow & -336x+392x& = & 35+80 \\\Leftrightarrow & \color{red}{56} x& = & 115 \\\Leftrightarrow & x = \frac{115}{56} & & \\ & V = \left\{ \frac{115}{56} \right\} & \\\end{align}\)
  9. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (-2x+\frac{2}{5})& = & -9x+\frac{2}{3} \\\Leftrightarrow & -8x+\frac{8}{5}& = & -9x+\frac{2}{3} \\ & & & \text{kgv van noemers 5 en 3 is 15} \\\Leftrightarrow & \color{blue}{15} .(\frac{-120}{ \color{blue}{15} }x+ \frac{24}{ \color{blue}{15} })& = & (\frac{-135}{ \color{blue}{15} }x+ \frac{10}{ \color{blue}{15} }). \color{blue}{15} \\\Leftrightarrow & -120x \color{red}{+24} & = & \color{red}{-135x} +10 \\\Leftrightarrow & -120x \color{red}{+24} \color{blue}{-24} \color{blue}{+135x} & = & \color{red}{-135x} +10 \color{blue}{+135x} \color{blue}{-24} \\\Leftrightarrow & -120x+135x& = & 10-24 \\\Leftrightarrow & \color{red}{15} x& = & -14 \\\Leftrightarrow & x = \frac{-14}{15} & & \\ & V = \left\{ \frac{-14}{15} \right\} & \\\end{align}\)
  10. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-5} (3x+\frac{4}{3})& = & -4x+\frac{5}{2} \\\Leftrightarrow & -15x-\frac{20}{3}& = & -4x+\frac{5}{2} \\ & & & \text{kgv van noemers 3 en 2 is 6} \\\Leftrightarrow & \color{blue}{6} .(\frac{-90}{ \color{blue}{6} }x- \frac{40}{ \color{blue}{6} })& = & (\frac{-24}{ \color{blue}{6} }x+ \frac{15}{ \color{blue}{6} }). \color{blue}{6} \\\Leftrightarrow & -90x \color{red}{-40} & = & \color{red}{-24x} +15 \\\Leftrightarrow & -90x \color{red}{-40} \color{blue}{+40} \color{blue}{+24x} & = & \color{red}{-24x} +15 \color{blue}{+24x} \color{blue}{+40} \\\Leftrightarrow & -90x+24x& = & 15+40 \\\Leftrightarrow & \color{red}{-66} x& = & 55 \\\Leftrightarrow & x = \frac{55}{-66} & & \\\Leftrightarrow & x = \frac{-5}{6} & & \\ & V = \left\{ \frac{-5}{6} \right\} & \\\end{align}\)
  11. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (5x-\frac{3}{11})& = & -9x+\frac{2}{5} \\\Leftrightarrow & -35x+\frac{21}{11}& = & -9x+\frac{2}{5} \\ & & & \text{kgv van noemers 11 en 5 is 55} \\\Leftrightarrow & \color{blue}{55} .(\frac{-1925}{ \color{blue}{55} }x+ \frac{105}{ \color{blue}{55} })& = & (\frac{-495}{ \color{blue}{55} }x+ \frac{22}{ \color{blue}{55} }). \color{blue}{55} \\\Leftrightarrow & -1925x \color{red}{+105} & = & \color{red}{-495x} +22 \\\Leftrightarrow & -1925x \color{red}{+105} \color{blue}{-105} \color{blue}{+495x} & = & \color{red}{-495x} +22 \color{blue}{+495x} \color{blue}{-105} \\\Leftrightarrow & -1925x+495x& = & 22-105 \\\Leftrightarrow & \color{red}{-1430} x& = & -83 \\\Leftrightarrow & x = \frac{-83}{-1430} & & \\\Leftrightarrow & x = \frac{83}{1430} & & \\ & V = \left\{ \frac{83}{1430} \right\} & \\\end{align}\)
  12. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (4x-\frac{4}{3})& = & -7x+\frac{6}{5} \\\Leftrightarrow & 20x-\frac{20}{3}& = & -7x+\frac{6}{5} \\ & & & \text{kgv van noemers 3 en 5 is 15} \\\Leftrightarrow & \color{blue}{15} .(\frac{300}{ \color{blue}{15} }x- \frac{100}{ \color{blue}{15} })& = & (\frac{-105}{ \color{blue}{15} }x+ \frac{18}{ \color{blue}{15} }). \color{blue}{15} \\\Leftrightarrow & 300x \color{red}{-100} & = & \color{red}{-105x} +18 \\\Leftrightarrow & 300x \color{red}{-100} \color{blue}{+100} \color{blue}{+105x} & = & \color{red}{-105x} +18 \color{blue}{+105x} \color{blue}{+100} \\\Leftrightarrow & 300x+105x& = & 18+100 \\\Leftrightarrow & \color{red}{405} x& = & 118 \\\Leftrightarrow & x = \frac{118}{405} & & \\ & V = \left\{ \frac{118}{405} \right\} & \\\end{align}\)
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