Vgln. eerste graad (reeks 5)

Hoofdmenu Eentje per keer 

Alles samen. Gebruik stappenplan en ZRM!

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

Alles samen. Gebruik stappenplan en ZRM!

Verbetersleutel

  1. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-6} (2x+\frac{3}{11})& = & 5x+\frac{6}{11} \\\Leftrightarrow & -12x-\frac{18}{11}& = & 5x+\frac{6}{11} \\ & & & \text{kgv van noemers 11 en 11 is 11} \\\Leftrightarrow & \color{blue}{11} .(\frac{-132}{ \color{blue}{11} }x- \frac{18}{ \color{blue}{11} })& = & (\frac{55}{ \color{blue}{11} }x+ \frac{6}{ \color{blue}{11} }). \color{blue}{11} \\\Leftrightarrow & -132x \color{red}{-18} & = & \color{red}{55x} +6 \\\Leftrightarrow & -132x \color{red}{-18} \color{blue}{+18} \color{blue}{-55x} & = & \color{red}{55x} +6 \color{blue}{-55x} \color{blue}{+18} \\\Leftrightarrow & -132x-55x& = & 6+18 \\\Leftrightarrow & \color{red}{-187} x& = & 24 \\\Leftrightarrow & x = \frac{24}{-187} & & \\\Leftrightarrow & x = \frac{-24}{187} & & \\ & V = \left\{ \frac{-24}{187} \right\} & \\\end{align}\)
  2. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (-4x+\frac{3}{7})& = & 9x+\frac{6}{11} \\\Leftrightarrow & 8x-\frac{6}{7}& = & 9x+\frac{6}{11} \\ & & & \text{kgv van noemers 7 en 11 is 77} \\\Leftrightarrow & \color{blue}{77} .(\frac{616}{ \color{blue}{77} }x- \frac{66}{ \color{blue}{77} })& = & (\frac{693}{ \color{blue}{77} }x+ \frac{42}{ \color{blue}{77} }). \color{blue}{77} \\\Leftrightarrow & 616x \color{red}{-66} & = & \color{red}{693x} +42 \\\Leftrightarrow & 616x \color{red}{-66} \color{blue}{+66} \color{blue}{-693x} & = & \color{red}{693x} +42 \color{blue}{-693x} \color{blue}{+66} \\\Leftrightarrow & 616x-693x& = & 42+66 \\\Leftrightarrow & \color{red}{-77} x& = & 108 \\\Leftrightarrow & x = \frac{108}{-77} & & \\\Leftrightarrow & x = \frac{-108}{77} & & \\ & V = \left\{ \frac{-108}{77} \right\} & \\\end{align}\)
  3. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-3} (5x-\frac{3}{10})& = & -4x+\frac{8}{3} \\\Leftrightarrow & -15x+\frac{9}{10}& = & -4x+\frac{8}{3} \\ & & & \text{kgv van noemers 10 en 3 is 30} \\\Leftrightarrow & \color{blue}{30} .(\frac{-450}{ \color{blue}{30} }x+ \frac{27}{ \color{blue}{30} })& = & (\frac{-120}{ \color{blue}{30} }x+ \frac{80}{ \color{blue}{30} }). \color{blue}{30} \\\Leftrightarrow & -450x \color{red}{+27} & = & \color{red}{-120x} +80 \\\Leftrightarrow & -450x \color{red}{+27} \color{blue}{-27} \color{blue}{+120x} & = & \color{red}{-120x} +80 \color{blue}{+120x} \color{blue}{-27} \\\Leftrightarrow & -450x+120x& = & 80-27 \\\Leftrightarrow & \color{red}{-330} x& = & 53 \\\Leftrightarrow & x = \frac{53}{-330} & & \\\Leftrightarrow & x = \frac{-53}{330} & & \\ & V = \left\{ \frac{-53}{330} \right\} & \\\end{align}\)
  4. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (-3x-\frac{4}{9})& = & -5x+\frac{3}{10} \\\Leftrightarrow & -12x-\frac{16}{9}& = & -5x+\frac{3}{10} \\ & & & \text{kgv van noemers 9 en 10 is 90} \\\Leftrightarrow & \color{blue}{90} .(\frac{-1080}{ \color{blue}{90} }x- \frac{160}{ \color{blue}{90} })& = & (\frac{-450}{ \color{blue}{90} }x+ \frac{27}{ \color{blue}{90} }). \color{blue}{90} \\\Leftrightarrow & -1080x \color{red}{-160} & = & \color{red}{-450x} +27 \\\Leftrightarrow & -1080x \color{red}{-160} \color{blue}{+160} \color{blue}{+450x} & = & \color{red}{-450x} +27 \color{blue}{+450x} \color{blue}{+160} \\\Leftrightarrow & -1080x+450x& = & 27+160 \\\Leftrightarrow & \color{red}{-630} x& = & 187 \\\Leftrightarrow & x = \frac{187}{-630} & & \\\Leftrightarrow & x = \frac{-187}{630} & & \\ & V = \left\{ \frac{-187}{630} \right\} & \\\end{align}\)
  5. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-3} (3x+\frac{5}{4})& = & -5x+\frac{8}{7} \\\Leftrightarrow & -9x-\frac{15}{4}& = & -5x+\frac{8}{7} \\ & & & \text{kgv van noemers 4 en 7 is 28} \\\Leftrightarrow & \color{blue}{28} .(\frac{-252}{ \color{blue}{28} }x- \frac{105}{ \color{blue}{28} })& = & (\frac{-140}{ \color{blue}{28} }x+ \frac{32}{ \color{blue}{28} }). \color{blue}{28} \\\Leftrightarrow & -252x \color{red}{-105} & = & \color{red}{-140x} +32 \\\Leftrightarrow & -252x \color{red}{-105} \color{blue}{+105} \color{blue}{+140x} & = & \color{red}{-140x} +32 \color{blue}{+140x} \color{blue}{+105} \\\Leftrightarrow & -252x+140x& = & 32+105 \\\Leftrightarrow & \color{red}{-112} x& = & 137 \\\Leftrightarrow & x = \frac{137}{-112} & & \\\Leftrightarrow & x = \frac{-137}{112} & & \\ & V = \left\{ \frac{-137}{112} \right\} & \\\end{align}\)
  6. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (-3x+\frac{3}{7})& = & -5x+\frac{4}{7} \\\Leftrightarrow & 6x-\frac{6}{7}& = & -5x+\frac{4}{7} \\ & & & \text{kgv van noemers 7 en 7 is 7} \\\Leftrightarrow & \color{blue}{7} .(\frac{42}{ \color{blue}{7} }x- \frac{6}{ \color{blue}{7} })& = & (\frac{-35}{ \color{blue}{7} }x+ \frac{4}{ \color{blue}{7} }). \color{blue}{7} \\\Leftrightarrow & 42x \color{red}{-6} & = & \color{red}{-35x} +4 \\\Leftrightarrow & 42x \color{red}{-6} \color{blue}{+6} \color{blue}{+35x} & = & \color{red}{-35x} +4 \color{blue}{+35x} \color{blue}{+6} \\\Leftrightarrow & 42x+35x& = & 4+6 \\\Leftrightarrow & \color{red}{77} x& = & 10 \\\Leftrightarrow & x = \frac{10}{77} & & \\ & V = \left\{ \frac{10}{77} \right\} & \\\end{align}\)
  7. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (-2x+\frac{2}{3})& = & 3x+\frac{4}{3} \\\Leftrightarrow & 14x-\frac{14}{3}& = & 3x+\frac{4}{3} \\ & & & \text{kgv van noemers 3 en 3 is 3} \\\Leftrightarrow & \color{blue}{3} .(\frac{42}{ \color{blue}{3} }x- \frac{14}{ \color{blue}{3} })& = & (\frac{9}{ \color{blue}{3} }x+ \frac{4}{ \color{blue}{3} }). \color{blue}{3} \\\Leftrightarrow & 42x \color{red}{-14} & = & \color{red}{9x} +4 \\\Leftrightarrow & 42x \color{red}{-14} \color{blue}{+14} \color{blue}{-9x} & = & \color{red}{9x} +4 \color{blue}{-9x} \color{blue}{+14} \\\Leftrightarrow & 42x-9x& = & 4+14 \\\Leftrightarrow & \color{red}{33} x& = & 18 \\\Leftrightarrow & x = \frac{18}{33} & & \\\Leftrightarrow & x = \frac{6}{11} & & \\ & V = \left\{ \frac{6}{11} \right\} & \\\end{align}\)
  8. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-5} (4x+\frac{4}{3})& = & -9x+\frac{7}{11} \\\Leftrightarrow & -20x-\frac{20}{3}& = & -9x+\frac{7}{11} \\ & & & \text{kgv van noemers 3 en 11 is 33} \\\Leftrightarrow & \color{blue}{33} .(\frac{-660}{ \color{blue}{33} }x- \frac{220}{ \color{blue}{33} })& = & (\frac{-297}{ \color{blue}{33} }x+ \frac{21}{ \color{blue}{33} }). \color{blue}{33} \\\Leftrightarrow & -660x \color{red}{-220} & = & \color{red}{-297x} +21 \\\Leftrightarrow & -660x \color{red}{-220} \color{blue}{+220} \color{blue}{+297x} & = & \color{red}{-297x} +21 \color{blue}{+297x} \color{blue}{+220} \\\Leftrightarrow & -660x+297x& = & 21+220 \\\Leftrightarrow & \color{red}{-363} x& = & 241 \\\Leftrightarrow & x = \frac{241}{-363} & & \\\Leftrightarrow & x = \frac{-241}{363} & & \\ & V = \left\{ \frac{-241}{363} \right\} & \\\end{align}\)
  9. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-4} (-4x+\frac{3}{5})& = & 5x+\frac{3}{2} \\\Leftrightarrow & 16x-\frac{12}{5}& = & 5x+\frac{3}{2} \\ & & & \text{kgv van noemers 5 en 2 is 10} \\\Leftrightarrow & \color{blue}{10} .(\frac{160}{ \color{blue}{10} }x- \frac{24}{ \color{blue}{10} })& = & (\frac{50}{ \color{blue}{10} }x+ \frac{15}{ \color{blue}{10} }). \color{blue}{10} \\\Leftrightarrow & 160x \color{red}{-24} & = & \color{red}{50x} +15 \\\Leftrightarrow & 160x \color{red}{-24} \color{blue}{+24} \color{blue}{-50x} & = & \color{red}{50x} +15 \color{blue}{-50x} \color{blue}{+24} \\\Leftrightarrow & 160x-50x& = & 15+24 \\\Leftrightarrow & \color{red}{110} x& = & 39 \\\Leftrightarrow & x = \frac{39}{110} & & \\ & V = \left\{ \frac{39}{110} \right\} & \\\end{align}\)
  10. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-6} (2x+\frac{2}{5})& = & -5x+\frac{7}{9} \\\Leftrightarrow & -12x-\frac{12}{5}& = & -5x+\frac{7}{9} \\ & & & \text{kgv van noemers 5 en 9 is 45} \\\Leftrightarrow & \color{blue}{45} .(\frac{-540}{ \color{blue}{45} }x- \frac{108}{ \color{blue}{45} })& = & (\frac{-225}{ \color{blue}{45} }x+ \frac{35}{ \color{blue}{45} }). \color{blue}{45} \\\Leftrightarrow & -540x \color{red}{-108} & = & \color{red}{-225x} +35 \\\Leftrightarrow & -540x \color{red}{-108} \color{blue}{+108} \color{blue}{+225x} & = & \color{red}{-225x} +35 \color{blue}{+225x} \color{blue}{+108} \\\Leftrightarrow & -540x+225x& = & 35+108 \\\Leftrightarrow & \color{red}{-315} x& = & 143 \\\Leftrightarrow & x = \frac{143}{-315} & & \\\Leftrightarrow & x = \frac{-143}{315} & & \\ & V = \left\{ \frac{-143}{315} \right\} & \\\end{align}\)
  11. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-6} (5x+\frac{2}{5})& = & -7x+\frac{4}{9} \\\Leftrightarrow & -30x-\frac{12}{5}& = & -7x+\frac{4}{9} \\ & & & \text{kgv van noemers 5 en 9 is 45} \\\Leftrightarrow & \color{blue}{45} .(\frac{-1350}{ \color{blue}{45} }x- \frac{108}{ \color{blue}{45} })& = & (\frac{-315}{ \color{blue}{45} }x+ \frac{20}{ \color{blue}{45} }). \color{blue}{45} \\\Leftrightarrow & -1350x \color{red}{-108} & = & \color{red}{-315x} +20 \\\Leftrightarrow & -1350x \color{red}{-108} \color{blue}{+108} \color{blue}{+315x} & = & \color{red}{-315x} +20 \color{blue}{+315x} \color{blue}{+108} \\\Leftrightarrow & -1350x+315x& = & 20+108 \\\Leftrightarrow & \color{red}{-1035} x& = & 128 \\\Leftrightarrow & x = \frac{128}{-1035} & & \\\Leftrightarrow & x = \frac{-128}{1035} & & \\ & V = \left\{ \frac{-128}{1035} \right\} & \\\end{align}\)
  12. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{2} (-3x-\frac{3}{5})& = & 7x+\frac{10}{9} \\\Leftrightarrow & -6x-\frac{6}{5}& = & 7x+\frac{10}{9} \\ & & & \text{kgv van noemers 5 en 9 is 45} \\\Leftrightarrow & \color{blue}{45} .(\frac{-270}{ \color{blue}{45} }x- \frac{54}{ \color{blue}{45} })& = & (\frac{315}{ \color{blue}{45} }x+ \frac{50}{ \color{blue}{45} }). \color{blue}{45} \\\Leftrightarrow & -270x \color{red}{-54} & = & \color{red}{315x} +50 \\\Leftrightarrow & -270x \color{red}{-54} \color{blue}{+54} \color{blue}{-315x} & = & \color{red}{315x} +50 \color{blue}{-315x} \color{blue}{+54} \\\Leftrightarrow & -270x-315x& = & 50+54 \\\Leftrightarrow & \color{red}{-585} x& = & 104 \\\Leftrightarrow & x = \frac{104}{-585} & & \\\Leftrightarrow & x = \frac{-8}{45} & & \\ & V = \left\{ \frac{-8}{45} \right\} & \\\end{align}\)
Oefeningengenerator wiskundeoefeningen.be 2026-09-24 09:02:17
Een site van Busleyden Atheneum Mechelen