Vgln. eerste graad (reeks 5)

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Alles samen. Gebruik stappenplan en ZRM!

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

Alles samen. Gebruik stappenplan en ZRM!

Verbetersleutel

  1. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{2} (5x+\frac{3}{5})& = & -7x+\frac{8}{3} \\\Leftrightarrow & 10x+\frac{6}{5}& = & -7x+\frac{8}{3} \\ & & & \text{kgv van noemers 5 en 3 is 15} \\\Leftrightarrow & \color{blue}{15} .(\frac{150}{ \color{blue}{15} }x+ \frac{18}{ \color{blue}{15} })& = & (\frac{-105}{ \color{blue}{15} }x+ \frac{40}{ \color{blue}{15} }). \color{blue}{15} \\\Leftrightarrow & 150x \color{red}{+18} & = & \color{red}{-105x} +40 \\\Leftrightarrow & 150x \color{red}{+18} \color{blue}{-18} \color{blue}{+105x} & = & \color{red}{-105x} +40 \color{blue}{+105x} \color{blue}{-18} \\\Leftrightarrow & 150x+105x& = & 40-18 \\\Leftrightarrow & \color{red}{255} x& = & 22 \\\Leftrightarrow & x = \frac{22}{255} & & \\ & V = \left\{ \frac{22}{255} \right\} & \\\end{align}\)
  2. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (2x+\frac{3}{11})& = & 7x+\frac{2}{11} \\\Leftrightarrow & 8x+\frac{12}{11}& = & 7x+\frac{2}{11} \\ & & & \text{kgv van noemers 11 en 11 is 11} \\\Leftrightarrow & \color{blue}{11} .(\frac{88}{ \color{blue}{11} }x+ \frac{12}{ \color{blue}{11} })& = & (\frac{77}{ \color{blue}{11} }x+ \frac{2}{ \color{blue}{11} }). \color{blue}{11} \\\Leftrightarrow & 88x \color{red}{+12} & = & \color{red}{77x} +2 \\\Leftrightarrow & 88x \color{red}{+12} \color{blue}{-12} \color{blue}{-77x} & = & \color{red}{77x} +2 \color{blue}{-77x} \color{blue}{-12} \\\Leftrightarrow & 88x-77x& = & 2-12 \\\Leftrightarrow & \color{red}{11} x& = & -10 \\\Leftrightarrow & x = \frac{-10}{11} & & \\ & V = \left\{ \frac{-10}{11} \right\} & \\\end{align}\)
  3. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (4x+\frac{3}{8})& = & -9x+\frac{9}{2} \\\Leftrightarrow & -28x-\frac{21}{8}& = & -9x+\frac{9}{2} \\ & & & \text{kgv van noemers 8 en 2 is 8} \\\Leftrightarrow & \color{blue}{8} .(\frac{-224}{ \color{blue}{8} }x- \frac{21}{ \color{blue}{8} })& = & (\frac{-72}{ \color{blue}{8} }x+ \frac{36}{ \color{blue}{8} }). \color{blue}{8} \\\Leftrightarrow & -224x \color{red}{-21} & = & \color{red}{-72x} +36 \\\Leftrightarrow & -224x \color{red}{-21} \color{blue}{+21} \color{blue}{+72x} & = & \color{red}{-72x} +36 \color{blue}{+72x} \color{blue}{+21} \\\Leftrightarrow & -224x+72x& = & 36+21 \\\Leftrightarrow & \color{red}{-152} x& = & 57 \\\Leftrightarrow & x = \frac{57}{-152} & & \\\Leftrightarrow & x = \frac{-3}{8} & & \\ & V = \left\{ \frac{-3}{8} \right\} & \\\end{align}\)
  4. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{6} (-4x+\frac{2}{5})& = & -5x+\frac{9}{2} \\\Leftrightarrow & -24x+\frac{12}{5}& = & -5x+\frac{9}{2} \\ & & & \text{kgv van noemers 5 en 2 is 10} \\\Leftrightarrow & \color{blue}{10} .(\frac{-240}{ \color{blue}{10} }x+ \frac{24}{ \color{blue}{10} })& = & (\frac{-50}{ \color{blue}{10} }x+ \frac{45}{ \color{blue}{10} }). \color{blue}{10} \\\Leftrightarrow & -240x \color{red}{+24} & = & \color{red}{-50x} +45 \\\Leftrightarrow & -240x \color{red}{+24} \color{blue}{-24} \color{blue}{+50x} & = & \color{red}{-50x} +45 \color{blue}{+50x} \color{blue}{-24} \\\Leftrightarrow & -240x+50x& = & 45-24 \\\Leftrightarrow & \color{red}{-190} x& = & 21 \\\Leftrightarrow & x = \frac{21}{-190} & & \\\Leftrightarrow & x = \frac{-21}{190} & & \\ & V = \left\{ \frac{-21}{190} \right\} & \\\end{align}\)
  5. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{7} (5x-\frac{4}{11})& = & -2x+\frac{7}{10} \\\Leftrightarrow & 35x-\frac{28}{11}& = & -2x+\frac{7}{10} \\ & & & \text{kgv van noemers 11 en 10 is 110} \\\Leftrightarrow & \color{blue}{110} .(\frac{3850}{ \color{blue}{110} }x- \frac{280}{ \color{blue}{110} })& = & (\frac{-220}{ \color{blue}{110} }x+ \frac{77}{ \color{blue}{110} }). \color{blue}{110} \\\Leftrightarrow & 3850x \color{red}{-280} & = & \color{red}{-220x} +77 \\\Leftrightarrow & 3850x \color{red}{-280} \color{blue}{+280} \color{blue}{+220x} & = & \color{red}{-220x} +77 \color{blue}{+220x} \color{blue}{+280} \\\Leftrightarrow & 3850x+220x& = & 77+280 \\\Leftrightarrow & \color{red}{4070} x& = & 357 \\\Leftrightarrow & x = \frac{357}{4070} & & \\ & V = \left\{ \frac{357}{4070} \right\} & \\\end{align}\)
  6. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{7} (3x-\frac{3}{10})& = & 8x+\frac{4}{11} \\\Leftrightarrow & 21x-\frac{21}{10}& = & 8x+\frac{4}{11} \\ & & & \text{kgv van noemers 10 en 11 is 110} \\\Leftrightarrow & \color{blue}{110} .(\frac{2310}{ \color{blue}{110} }x- \frac{231}{ \color{blue}{110} })& = & (\frac{880}{ \color{blue}{110} }x+ \frac{40}{ \color{blue}{110} }). \color{blue}{110} \\\Leftrightarrow & 2310x \color{red}{-231} & = & \color{red}{880x} +40 \\\Leftrightarrow & 2310x \color{red}{-231} \color{blue}{+231} \color{blue}{-880x} & = & \color{red}{880x} +40 \color{blue}{-880x} \color{blue}{+231} \\\Leftrightarrow & 2310x-880x& = & 40+231 \\\Leftrightarrow & \color{red}{1430} x& = & 271 \\\Leftrightarrow & x = \frac{271}{1430} & & \\ & V = \left\{ \frac{271}{1430} \right\} & \\\end{align}\)
  7. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (4x-\frac{2}{5})& = & -7x+\frac{4}{7} \\\Leftrightarrow & 16x-\frac{8}{5}& = & -7x+\frac{4}{7} \\ & & & \text{kgv van noemers 5 en 7 is 35} \\\Leftrightarrow & \color{blue}{35} .(\frac{560}{ \color{blue}{35} }x- \frac{56}{ \color{blue}{35} })& = & (\frac{-245}{ \color{blue}{35} }x+ \frac{20}{ \color{blue}{35} }). \color{blue}{35} \\\Leftrightarrow & 560x \color{red}{-56} & = & \color{red}{-245x} +20 \\\Leftrightarrow & 560x \color{red}{-56} \color{blue}{+56} \color{blue}{+245x} & = & \color{red}{-245x} +20 \color{blue}{+245x} \color{blue}{+56} \\\Leftrightarrow & 560x+245x& = & 20+56 \\\Leftrightarrow & \color{red}{805} x& = & 76 \\\Leftrightarrow & x = \frac{76}{805} & & \\ & V = \left\{ \frac{76}{805} \right\} & \\\end{align}\)
  8. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{2} (3x+\frac{5}{3})& = & -5x+\frac{5}{12} \\\Leftrightarrow & 6x+\frac{10}{3}& = & -5x+\frac{5}{12} \\ & & & \text{kgv van noemers 3 en 12 is 12} \\\Leftrightarrow & \color{blue}{12} .(\frac{72}{ \color{blue}{12} }x+ \frac{40}{ \color{blue}{12} })& = & (\frac{-60}{ \color{blue}{12} }x+ \frac{5}{ \color{blue}{12} }). \color{blue}{12} \\\Leftrightarrow & 72x \color{red}{+40} & = & \color{red}{-60x} +5 \\\Leftrightarrow & 72x \color{red}{+40} \color{blue}{-40} \color{blue}{+60x} & = & \color{red}{-60x} +5 \color{blue}{+60x} \color{blue}{-40} \\\Leftrightarrow & 72x+60x& = & 5-40 \\\Leftrightarrow & \color{red}{132} x& = & -35 \\\Leftrightarrow & x = \frac{-35}{132} & & \\ & V = \left\{ \frac{-35}{132} \right\} & \\\end{align}\)
  9. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (5x+\frac{5}{2})& = & -2x+\frac{7}{12} \\\Leftrightarrow & 15x+\frac{15}{2}& = & -2x+\frac{7}{12} \\ & & & \text{kgv van noemers 2 en 12 is 12} \\\Leftrightarrow & \color{blue}{12} .(\frac{180}{ \color{blue}{12} }x+ \frac{90}{ \color{blue}{12} })& = & (\frac{-24}{ \color{blue}{12} }x+ \frac{7}{ \color{blue}{12} }). \color{blue}{12} \\\Leftrightarrow & 180x \color{red}{+90} & = & \color{red}{-24x} +7 \\\Leftrightarrow & 180x \color{red}{+90} \color{blue}{-90} \color{blue}{+24x} & = & \color{red}{-24x} +7 \color{blue}{+24x} \color{blue}{-90} \\\Leftrightarrow & 180x+24x& = & 7-90 \\\Leftrightarrow & \color{red}{204} x& = & -83 \\\Leftrightarrow & x = \frac{-83}{204} & & \\ & V = \left\{ \frac{-83}{204} \right\} & \\\end{align}\)
  10. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-5} (4x-\frac{5}{8})& = & 7x+\frac{5}{4} \\\Leftrightarrow & -20x+\frac{25}{8}& = & 7x+\frac{5}{4} \\ & & & \text{kgv van noemers 8 en 4 is 8} \\\Leftrightarrow & \color{blue}{8} .(\frac{-160}{ \color{blue}{8} }x+ \frac{25}{ \color{blue}{8} })& = & (\frac{56}{ \color{blue}{8} }x+ \frac{10}{ \color{blue}{8} }). \color{blue}{8} \\\Leftrightarrow & -160x \color{red}{+25} & = & \color{red}{56x} +10 \\\Leftrightarrow & -160x \color{red}{+25} \color{blue}{-25} \color{blue}{-56x} & = & \color{red}{56x} +10 \color{blue}{-56x} \color{blue}{-25} \\\Leftrightarrow & -160x-56x& = & 10-25 \\\Leftrightarrow & \color{red}{-216} x& = & -15 \\\Leftrightarrow & x = \frac{-15}{-216} & & \\\Leftrightarrow & x = \frac{5}{72} & & \\ & V = \left\{ \frac{5}{72} \right\} & \\\end{align}\)
  11. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (5x-\frac{2}{5})& = & 7x+\frac{7}{11} \\\Leftrightarrow & 15x-\frac{6}{5}& = & 7x+\frac{7}{11} \\ & & & \text{kgv van noemers 5 en 11 is 55} \\\Leftrightarrow & \color{blue}{55} .(\frac{825}{ \color{blue}{55} }x- \frac{66}{ \color{blue}{55} })& = & (\frac{385}{ \color{blue}{55} }x+ \frac{35}{ \color{blue}{55} }). \color{blue}{55} \\\Leftrightarrow & 825x \color{red}{-66} & = & \color{red}{385x} +35 \\\Leftrightarrow & 825x \color{red}{-66} \color{blue}{+66} \color{blue}{-385x} & = & \color{red}{385x} +35 \color{blue}{-385x} \color{blue}{+66} \\\Leftrightarrow & 825x-385x& = & 35+66 \\\Leftrightarrow & \color{red}{440} x& = & 101 \\\Leftrightarrow & x = \frac{101}{440} & & \\ & V = \left\{ \frac{101}{440} \right\} & \\\end{align}\)
  12. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (4x+\frac{2}{11})& = & 7x+\frac{10}{3} \\\Leftrightarrow & 12x+\frac{6}{11}& = & 7x+\frac{10}{3} \\ & & & \text{kgv van noemers 11 en 3 is 33} \\\Leftrightarrow & \color{blue}{33} .(\frac{396}{ \color{blue}{33} }x+ \frac{18}{ \color{blue}{33} })& = & (\frac{231}{ \color{blue}{33} }x+ \frac{110}{ \color{blue}{33} }). \color{blue}{33} \\\Leftrightarrow & 396x \color{red}{+18} & = & \color{red}{231x} +110 \\\Leftrightarrow & 396x \color{red}{+18} \color{blue}{-18} \color{blue}{-231x} & = & \color{red}{231x} +110 \color{blue}{-231x} \color{blue}{-18} \\\Leftrightarrow & 396x-231x& = & 110-18 \\\Leftrightarrow & \color{red}{165} x& = & 92 \\\Leftrightarrow & x = \frac{92}{165} & & \\ & V = \left\{ \frac{92}{165} \right\} & \\\end{align}\)
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