Vgln. eerste graad (reeks 5)

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Alles samen. Gebruik stappenplan en ZRM!

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

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Alles samen. Gebruik stappenplan en ZRM!

Verbetersleutel

1. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (5x-\frac{2}{3})& = & -3x+\frac{6}{5} \\\Leftrightarrow & -35x+\frac{14}{3}& = & -3x+\frac{6}{5} \\ & & & \text{kgv van noemers 3 en 5 is 15} \\\Leftrightarrow & \color{blue}{15} .(\frac{-525}{ \color{blue}{15} }x+ \frac{70}{ \color{blue}{15} })& = & (\frac{-45}{ \color{blue}{15} }x+ \frac{18}{ \color{blue}{15} }). \color{blue}{15} \\\Leftrightarrow & -525x \color{red}{+70} & = & \color{red}{-45x} +18 \\\Leftrightarrow & -525x \color{red}{+70} \color{blue}{-70} \color{blue}{+45x} & = & \color{red}{-45x} +18 \color{blue}{+45x} \color{blue}{-70} \\\Leftrightarrow & -525x+45x& = & 18-70 \\\Leftrightarrow & \color{red}{-480} x& = & -52 \\\Leftrightarrow & x = \frac{-52}{-480} & & \\\Leftrightarrow & x = \frac{13}{120} & & \\ & V = \left\{ \frac{13}{120} \right\} & \\\end{align}\)
2. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{7} (5x-\frac{2}{11})& = & -8x+\frac{2}{11} \\\Leftrightarrow & 35x-\frac{14}{11}& = & -8x+\frac{2}{11} \\ & & & \text{kgv van noemers 11 en 11 is 11} \\\Leftrightarrow & \color{blue}{11} .(\frac{385}{ \color{blue}{11} }x- \frac{14}{ \color{blue}{11} })& = & (\frac{-88}{ \color{blue}{11} }x+ \frac{2}{ \color{blue}{11} }). \color{blue}{11} \\\Leftrightarrow & 385x \color{red}{-14} & = & \color{red}{-88x} +2 \\\Leftrightarrow & 385x \color{red}{-14} \color{blue}{+14} \color{blue}{+88x} & = & \color{red}{-88x} +2 \color{blue}{+88x} \color{blue}{+14} \\\Leftrightarrow & 385x+88x& = & 2+14 \\\Leftrightarrow & \color{red}{473} x& = & 16 \\\Leftrightarrow & x = \frac{16}{473} & & \\ & V = \left\{ \frac{16}{473} \right\} & \\\end{align}\)
3. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (-2x+\frac{5}{9})& = & -9x+\frac{6}{7} \\\Leftrightarrow & -8x+\frac{20}{9}& = & -9x+\frac{6}{7} \\ & & & \text{kgv van noemers 9 en 7 is 63} \\\Leftrightarrow & \color{blue}{63} .(\frac{-504}{ \color{blue}{63} }x+ \frac{140}{ \color{blue}{63} })& = & (\frac{-567}{ \color{blue}{63} }x+ \frac{54}{ \color{blue}{63} }). \color{blue}{63} \\\Leftrightarrow & -504x \color{red}{+140} & = & \color{red}{-567x} +54 \\\Leftrightarrow & -504x \color{red}{+140} \color{blue}{-140} \color{blue}{+567x} & = & \color{red}{-567x} +54 \color{blue}{+567x} \color{blue}{-140} \\\Leftrightarrow & -504x+567x& = & 54-140 \\\Leftrightarrow & \color{red}{63} x& = & -86 \\\Leftrightarrow & x = \frac{-86}{63} & & \\ & V = \left\{ \frac{-86}{63} \right\} & \\\end{align}\)
4. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (4x+\frac{4}{9})& = & 3x+\frac{7}{11} \\\Leftrightarrow & -8x-\frac{8}{9}& = & 3x+\frac{7}{11} \\ & & & \text{kgv van noemers 9 en 11 is 99} \\\Leftrightarrow & \color{blue}{99} .(\frac{-792}{ \color{blue}{99} }x- \frac{88}{ \color{blue}{99} })& = & (\frac{297}{ \color{blue}{99} }x+ \frac{63}{ \color{blue}{99} }). \color{blue}{99} \\\Leftrightarrow & -792x \color{red}{-88} & = & \color{red}{297x} +63 \\\Leftrightarrow & -792x \color{red}{-88} \color{blue}{+88} \color{blue}{-297x} & = & \color{red}{297x} +63 \color{blue}{-297x} \color{blue}{+88} \\\Leftrightarrow & -792x-297x& = & 63+88 \\\Leftrightarrow & \color{red}{-1089} x& = & 151 \\\Leftrightarrow & x = \frac{151}{-1089} & & \\\Leftrightarrow & x = \frac{-151}{1089} & & \\ & V = \left\{ \frac{-151}{1089} \right\} & \\\end{align}\)
5. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{2} (3x-\frac{5}{7})& = & -5x+\frac{9}{4} \\\Leftrightarrow & 6x-\frac{10}{7}& = & -5x+\frac{9}{4} \\ & & & \text{kgv van noemers 7 en 4 is 28} \\\Leftrightarrow & \color{blue}{28} .(\frac{168}{ \color{blue}{28} }x- \frac{40}{ \color{blue}{28} })& = & (\frac{-140}{ \color{blue}{28} }x+ \frac{63}{ \color{blue}{28} }). \color{blue}{28} \\\Leftrightarrow & 168x \color{red}{-40} & = & \color{red}{-140x} +63 \\\Leftrightarrow & 168x \color{red}{-40} \color{blue}{+40} \color{blue}{+140x} & = & \color{red}{-140x} +63 \color{blue}{+140x} \color{blue}{+40} \\\Leftrightarrow & 168x+140x& = & 63+40 \\\Leftrightarrow & \color{red}{308} x& = & 103 \\\Leftrightarrow & x = \frac{103}{308} & & \\ & V = \left\{ \frac{103}{308} \right\} & \\\end{align}\)
6. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (-4x-\frac{3}{8})& = & -9x+\frac{7}{6} \\\Leftrightarrow & 28x+\frac{21}{8}& = & -9x+\frac{7}{6} \\ & & & \text{kgv van noemers 8 en 6 is 24} \\\Leftrightarrow & \color{blue}{24} .(\frac{672}{ \color{blue}{24} }x+ \frac{63}{ \color{blue}{24} })& = & (\frac{-216}{ \color{blue}{24} }x+ \frac{28}{ \color{blue}{24} }). \color{blue}{24} \\\Leftrightarrow & 672x \color{red}{+63} & = & \color{red}{-216x} +28 \\\Leftrightarrow & 672x \color{red}{+63} \color{blue}{-63} \color{blue}{+216x} & = & \color{red}{-216x} +28 \color{blue}{+216x} \color{blue}{-63} \\\Leftrightarrow & 672x+216x& = & 28-63 \\\Leftrightarrow & \color{red}{888} x& = & -35 \\\Leftrightarrow & x = \frac{-35}{888} & & \\ & V = \left\{ \frac{-35}{888} \right\} & \\\end{align}\)
7. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-3} (2x+\frac{2}{7})& = & -7x+\frac{9}{4} \\\Leftrightarrow & -6x-\frac{6}{7}& = & -7x+\frac{9}{4} \\ & & & \text{kgv van noemers 7 en 4 is 28} \\\Leftrightarrow & \color{blue}{28} .(\frac{-168}{ \color{blue}{28} }x- \frac{24}{ \color{blue}{28} })& = & (\frac{-196}{ \color{blue}{28} }x+ \frac{63}{ \color{blue}{28} }). \color{blue}{28} \\\Leftrightarrow & -168x \color{red}{-24} & = & \color{red}{-196x} +63 \\\Leftrightarrow & -168x \color{red}{-24} \color{blue}{+24} \color{blue}{+196x} & = & \color{red}{-196x} +63 \color{blue}{+196x} \color{blue}{+24} \\\Leftrightarrow & -168x+196x& = & 63+24 \\\Leftrightarrow & \color{red}{28} x& = & 87 \\\Leftrightarrow & x = \frac{87}{28} & & \\ & V = \left\{ \frac{87}{28} \right\} & \\\end{align}\)
8. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-4} (2x-\frac{2}{9})& = & -9x+\frac{3}{2} \\\Leftrightarrow & -8x+\frac{8}{9}& = & -9x+\frac{3}{2} \\ & & & \text{kgv van noemers 9 en 2 is 18} \\\Leftrightarrow & \color{blue}{18} .(\frac{-144}{ \color{blue}{18} }x+ \frac{16}{ \color{blue}{18} })& = & (\frac{-162}{ \color{blue}{18} }x+ \frac{27}{ \color{blue}{18} }). \color{blue}{18} \\\Leftrightarrow & -144x \color{red}{+16} & = & \color{red}{-162x} +27 \\\Leftrightarrow & -144x \color{red}{+16} \color{blue}{-16} \color{blue}{+162x} & = & \color{red}{-162x} +27 \color{blue}{+162x} \color{blue}{-16} \\\Leftrightarrow & -144x+162x& = & 27-16 \\\Leftrightarrow & \color{red}{18} x& = & 11 \\\Leftrightarrow & x = \frac{11}{18} & & \\ & V = \left\{ \frac{11}{18} \right\} & \\\end{align}\)
9. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (-2x-\frac{3}{4})& = & 7x+\frac{7}{10} \\\Leftrightarrow & -10x-\frac{15}{4}& = & 7x+\frac{7}{10} \\ & & & \text{kgv van noemers 4 en 10 is 20} \\\Leftrightarrow & \color{blue}{20} .(\frac{-200}{ \color{blue}{20} }x- \frac{75}{ \color{blue}{20} })& = & (\frac{140}{ \color{blue}{20} }x+ \frac{14}{ \color{blue}{20} }). \color{blue}{20} \\\Leftrightarrow & -200x \color{red}{-75} & = & \color{red}{140x} +14 \\\Leftrightarrow & -200x \color{red}{-75} \color{blue}{+75} \color{blue}{-140x} & = & \color{red}{140x} +14 \color{blue}{-140x} \color{blue}{+75} \\\Leftrightarrow & -200x-140x& = & 14+75 \\\Leftrightarrow & \color{red}{-340} x& = & 89 \\\Leftrightarrow & x = \frac{89}{-340} & & \\\Leftrightarrow & x = \frac{-89}{340} & & \\ & V = \left\{ \frac{-89}{340} \right\} & \\\end{align}\)
10. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (3x-\frac{3}{7})& = & -7x+\frac{2}{11} \\\Leftrightarrow & -6x+\frac{6}{7}& = & -7x+\frac{2}{11} \\ & & & \text{kgv van noemers 7 en 11 is 77} \\\Leftrightarrow & \color{blue}{77} .(\frac{-462}{ \color{blue}{77} }x+ \frac{66}{ \color{blue}{77} })& = & (\frac{-539}{ \color{blue}{77} }x+ \frac{14}{ \color{blue}{77} }). \color{blue}{77} \\\Leftrightarrow & -462x \color{red}{+66} & = & \color{red}{-539x} +14 \\\Leftrightarrow & -462x \color{red}{+66} \color{blue}{-66} \color{blue}{+539x} & = & \color{red}{-539x} +14 \color{blue}{+539x} \color{blue}{-66} \\\Leftrightarrow & -462x+539x& = & 14-66 \\\Leftrightarrow & \color{red}{77} x& = & -52 \\\Leftrightarrow & x = \frac{-52}{77} & & \\ & V = \left\{ \frac{-52}{77} \right\} & \\\end{align}\)
11. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (-2x-\frac{1}{3})& = & -9x+\frac{3}{5} \\\Leftrightarrow & 14x+\frac{7}{3}& = & -9x+\frac{3}{5} \\ & & & \text{kgv van noemers 3 en 5 is 15} \\\Leftrightarrow & \color{blue}{15} .(\frac{210}{ \color{blue}{15} }x+ \frac{35}{ \color{blue}{15} })& = & (\frac{-135}{ \color{blue}{15} }x+ \frac{9}{ \color{blue}{15} }). \color{blue}{15} \\\Leftrightarrow & 210x \color{red}{+35} & = & \color{red}{-135x} +9 \\\Leftrightarrow & 210x \color{red}{+35} \color{blue}{-35} \color{blue}{+135x} & = & \color{red}{-135x} +9 \color{blue}{+135x} \color{blue}{-35} \\\Leftrightarrow & 210x+135x& = & 9-35 \\\Leftrightarrow & \color{red}{345} x& = & -26 \\\Leftrightarrow & x = \frac{-26}{345} & & \\ & V = \left\{ \frac{-26}{345} \right\} & \\\end{align}\)
12. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-6} (5x-\frac{4}{5})& = & -7x+\frac{7}{2} \\\Leftrightarrow & -30x+\frac{24}{5}& = & -7x+\frac{7}{2} \\ & & & \text{kgv van noemers 5 en 2 is 10} \\\Leftrightarrow & \color{blue}{10} .(\frac{-300}{ \color{blue}{10} }x+ \frac{48}{ \color{blue}{10} })& = & (\frac{-70}{ \color{blue}{10} }x+ \frac{35}{ \color{blue}{10} }). \color{blue}{10} \\\Leftrightarrow & -300x \color{red}{+48} & = & \color{red}{-70x} +35 \\\Leftrightarrow & -300x \color{red}{+48} \color{blue}{-48} \color{blue}{+70x} & = & \color{red}{-70x} +35 \color{blue}{+70x} \color{blue}{-48} \\\Leftrightarrow & -300x+70x& = & 35-48 \\\Leftrightarrow & \color{red}{-230} x& = & -13 \\\Leftrightarrow & x = \frac{-13}{-230} & & \\\Leftrightarrow & x = \frac{13}{230} & & \\ & V = \left\{ \frac{13}{230} \right\} & \\\end{align}\)