Vgln. eerste graad (reeks 5)

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Alles samen. Gebruik stappenplan en ZRM!

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

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Alles samen. Gebruik stappenplan en ZRM!

Verbetersleutel

1. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-4} (2x-\frac{3}{5})& = & -9x+\frac{7}{10} \\\Leftrightarrow & -8x+\frac{12}{5}& = & -9x+\frac{7}{10} \\ & & & \text{kgv van noemers 5 en 10 is 10} \\\Leftrightarrow & \color{blue}{10} .(\frac{-80}{ \color{blue}{10} }x+ \frac{24}{ \color{blue}{10} })& = & (\frac{-90}{ \color{blue}{10} }x+ \frac{7}{ \color{blue}{10} }). \color{blue}{10} \\\Leftrightarrow & -80x \color{red}{+24} & = & \color{red}{-90x} +7 \\\Leftrightarrow & -80x \color{red}{+24} \color{blue}{-24} \color{blue}{+90x} & = & \color{red}{-90x} +7 \color{blue}{+90x} \color{blue}{-24} \\\Leftrightarrow & -80x+90x& = & 7-24 \\\Leftrightarrow & \color{red}{10} x& = & -17 \\\Leftrightarrow & x = \frac{-17}{10} & & \\ & V = \left\{ \frac{-17}{10} \right\} & \\\end{align}\)
2. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (-2x-\frac{3}{10})& = & 7x+\frac{7}{9} \\\Leftrightarrow & -6x-\frac{9}{10}& = & 7x+\frac{7}{9} \\ & & & \text{kgv van noemers 10 en 9 is 90} \\\Leftrightarrow & \color{blue}{90} .(\frac{-540}{ \color{blue}{90} }x- \frac{81}{ \color{blue}{90} })& = & (\frac{630}{ \color{blue}{90} }x+ \frac{70}{ \color{blue}{90} }). \color{blue}{90} \\\Leftrightarrow & -540x \color{red}{-81} & = & \color{red}{630x} +70 \\\Leftrightarrow & -540x \color{red}{-81} \color{blue}{+81} \color{blue}{-630x} & = & \color{red}{630x} +70 \color{blue}{-630x} \color{blue}{+81} \\\Leftrightarrow & -540x-630x& = & 70+81 \\\Leftrightarrow & \color{red}{-1170} x& = & 151 \\\Leftrightarrow & x = \frac{151}{-1170} & & \\\Leftrightarrow & x = \frac{-151}{1170} & & \\ & V = \left\{ \frac{-151}{1170} \right\} & \\\end{align}\)
3. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-3} (2x-\frac{3}{10})& = & -7x+\frac{4}{9} \\\Leftrightarrow & -6x+\frac{9}{10}& = & -7x+\frac{4}{9} \\ & & & \text{kgv van noemers 10 en 9 is 90} \\\Leftrightarrow & \color{blue}{90} .(\frac{-540}{ \color{blue}{90} }x+ \frac{81}{ \color{blue}{90} })& = & (\frac{-630}{ \color{blue}{90} }x+ \frac{40}{ \color{blue}{90} }). \color{blue}{90} \\\Leftrightarrow & -540x \color{red}{+81} & = & \color{red}{-630x} +40 \\\Leftrightarrow & -540x \color{red}{+81} \color{blue}{-81} \color{blue}{+630x} & = & \color{red}{-630x} +40 \color{blue}{+630x} \color{blue}{-81} \\\Leftrightarrow & -540x+630x& = & 40-81 \\\Leftrightarrow & \color{red}{90} x& = & -41 \\\Leftrightarrow & x = \frac{-41}{90} & & \\ & V = \left\{ \frac{-41}{90} \right\} & \\\end{align}\)
4. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-6} (-3x-\frac{3}{11})& = & 5x+\frac{4}{3} \\\Leftrightarrow & 18x+\frac{18}{11}& = & 5x+\frac{4}{3} \\ & & & \text{kgv van noemers 11 en 3 is 33} \\\Leftrightarrow & \color{blue}{33} .(\frac{594}{ \color{blue}{33} }x+ \frac{54}{ \color{blue}{33} })& = & (\frac{165}{ \color{blue}{33} }x+ \frac{44}{ \color{blue}{33} }). \color{blue}{33} \\\Leftrightarrow & 594x \color{red}{+54} & = & \color{red}{165x} +44 \\\Leftrightarrow & 594x \color{red}{+54} \color{blue}{-54} \color{blue}{-165x} & = & \color{red}{165x} +44 \color{blue}{-165x} \color{blue}{-54} \\\Leftrightarrow & 594x-165x& = & 44-54 \\\Leftrightarrow & \color{red}{429} x& = & -10 \\\Leftrightarrow & x = \frac{-10}{429} & & \\ & V = \left\{ \frac{-10}{429} \right\} & \\\end{align}\)
5. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{6} (-2x-\frac{4}{5})& = & 5x+\frac{10}{9} \\\Leftrightarrow & -12x-\frac{24}{5}& = & 5x+\frac{10}{9} \\ & & & \text{kgv van noemers 5 en 9 is 45} \\\Leftrightarrow & \color{blue}{45} .(\frac{-540}{ \color{blue}{45} }x- \frac{216}{ \color{blue}{45} })& = & (\frac{225}{ \color{blue}{45} }x+ \frac{50}{ \color{blue}{45} }). \color{blue}{45} \\\Leftrightarrow & -540x \color{red}{-216} & = & \color{red}{225x} +50 \\\Leftrightarrow & -540x \color{red}{-216} \color{blue}{+216} \color{blue}{-225x} & = & \color{red}{225x} +50 \color{blue}{-225x} \color{blue}{+216} \\\Leftrightarrow & -540x-225x& = & 50+216 \\\Leftrightarrow & \color{red}{-765} x& = & 266 \\\Leftrightarrow & x = \frac{266}{-765} & & \\\Leftrightarrow & x = \frac{-266}{765} & & \\ & V = \left\{ \frac{-266}{765} \right\} & \\\end{align}\)
6. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (4x-\frac{3}{10})& = & -5x+\frac{2}{3} \\\Leftrightarrow & 12x-\frac{9}{10}& = & -5x+\frac{2}{3} \\ & & & \text{kgv van noemers 10 en 3 is 30} \\\Leftrightarrow & \color{blue}{30} .(\frac{360}{ \color{blue}{30} }x- \frac{27}{ \color{blue}{30} })& = & (\frac{-150}{ \color{blue}{30} }x+ \frac{20}{ \color{blue}{30} }). \color{blue}{30} \\\Leftrightarrow & 360x \color{red}{-27} & = & \color{red}{-150x} +20 \\\Leftrightarrow & 360x \color{red}{-27} \color{blue}{+27} \color{blue}{+150x} & = & \color{red}{-150x} +20 \color{blue}{+150x} \color{blue}{+27} \\\Leftrightarrow & 360x+150x& = & 20+27 \\\Leftrightarrow & \color{red}{510} x& = & 47 \\\Leftrightarrow & x = \frac{47}{510} & & \\ & V = \left\{ \frac{47}{510} \right\} & \\\end{align}\)
7. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{2} (2x+\frac{2}{3})& = & 9x+\frac{5}{9} \\\Leftrightarrow & 4x+\frac{4}{3}& = & 9x+\frac{5}{9} \\ & & & \text{kgv van noemers 3 en 9 is 9} \\\Leftrightarrow & \color{blue}{9} .(\frac{36}{ \color{blue}{9} }x+ \frac{12}{ \color{blue}{9} })& = & (\frac{81}{ \color{blue}{9} }x+ \frac{5}{ \color{blue}{9} }). \color{blue}{9} \\\Leftrightarrow & 36x \color{red}{+12} & = & \color{red}{81x} +5 \\\Leftrightarrow & 36x \color{red}{+12} \color{blue}{-12} \color{blue}{-81x} & = & \color{red}{81x} +5 \color{blue}{-81x} \color{blue}{-12} \\\Leftrightarrow & 36x-81x& = & 5-12 \\\Leftrightarrow & \color{red}{-45} x& = & -7 \\\Leftrightarrow & x = \frac{-7}{-45} & & \\\Leftrightarrow & x = \frac{7}{45} & & \\ & V = \left\{ \frac{7}{45} \right\} & \\\end{align}\)
8. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (2x-\frac{4}{3})& = & -5x+\frac{5}{6} \\\Leftrightarrow & 8x-\frac{16}{3}& = & -5x+\frac{5}{6} \\ & & & \text{kgv van noemers 3 en 6 is 6} \\\Leftrightarrow & \color{blue}{6} .(\frac{48}{ \color{blue}{6} }x- \frac{32}{ \color{blue}{6} })& = & (\frac{-30}{ \color{blue}{6} }x+ \frac{5}{ \color{blue}{6} }). \color{blue}{6} \\\Leftrightarrow & 48x \color{red}{-32} & = & \color{red}{-30x} +5 \\\Leftrightarrow & 48x \color{red}{-32} \color{blue}{+32} \color{blue}{+30x} & = & \color{red}{-30x} +5 \color{blue}{+30x} \color{blue}{+32} \\\Leftrightarrow & 48x+30x& = & 5+32 \\\Leftrightarrow & \color{red}{78} x& = & 37 \\\Leftrightarrow & x = \frac{37}{78} & & \\ & V = \left\{ \frac{37}{78} \right\} & \\\end{align}\)
9. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{3} (5x-\frac{5}{2})& = & -2x+\frac{7}{3} \\\Leftrightarrow & 15x-\frac{15}{2}& = & -2x+\frac{7}{3} \\ & & & \text{kgv van noemers 2 en 3 is 6} \\\Leftrightarrow & \color{blue}{6} .(\frac{90}{ \color{blue}{6} }x- \frac{45}{ \color{blue}{6} })& = & (\frac{-12}{ \color{blue}{6} }x+ \frac{14}{ \color{blue}{6} }). \color{blue}{6} \\\Leftrightarrow & 90x \color{red}{-45} & = & \color{red}{-12x} +14 \\\Leftrightarrow & 90x \color{red}{-45} \color{blue}{+45} \color{blue}{+12x} & = & \color{red}{-12x} +14 \color{blue}{+12x} \color{blue}{+45} \\\Leftrightarrow & 90x+12x& = & 14+45 \\\Leftrightarrow & \color{red}{102} x& = & 59 \\\Leftrightarrow & x = \frac{59}{102} & & \\ & V = \left\{ \frac{59}{102} \right\} & \\\end{align}\)
10. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (-5x+\frac{5}{11})& = & -3x+\frac{4}{3} \\\Leftrightarrow & -20x+\frac{20}{11}& = & -3x+\frac{4}{3} \\ & & & \text{kgv van noemers 11 en 3 is 33} \\\Leftrightarrow & \color{blue}{33} .(\frac{-660}{ \color{blue}{33} }x+ \frac{60}{ \color{blue}{33} })& = & (\frac{-99}{ \color{blue}{33} }x+ \frac{44}{ \color{blue}{33} }). \color{blue}{33} \\\Leftrightarrow & -660x \color{red}{+60} & = & \color{red}{-99x} +44 \\\Leftrightarrow & -660x \color{red}{+60} \color{blue}{-60} \color{blue}{+99x} & = & \color{red}{-99x} +44 \color{blue}{+99x} \color{blue}{-60} \\\Leftrightarrow & -660x+99x& = & 44-60 \\\Leftrightarrow & \color{red}{-561} x& = & -16 \\\Leftrightarrow & x = \frac{-16}{-561} & & \\\Leftrightarrow & x = \frac{16}{561} & & \\ & V = \left\{ \frac{16}{561} \right\} & \\\end{align}\)
11. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (-2x-\frac{5}{7})& = & 5x+\frac{7}{11} \\\Leftrightarrow & 4x+\frac{10}{7}& = & 5x+\frac{7}{11} \\ & & & \text{kgv van noemers 7 en 11 is 77} \\\Leftrightarrow & \color{blue}{77} .(\frac{308}{ \color{blue}{77} }x+ \frac{110}{ \color{blue}{77} })& = & (\frac{385}{ \color{blue}{77} }x+ \frac{49}{ \color{blue}{77} }). \color{blue}{77} \\\Leftrightarrow & 308x \color{red}{+110} & = & \color{red}{385x} +49 \\\Leftrightarrow & 308x \color{red}{+110} \color{blue}{-110} \color{blue}{-385x} & = & \color{red}{385x} +49 \color{blue}{-385x} \color{blue}{-110} \\\Leftrightarrow & 308x-385x& = & 49-110 \\\Leftrightarrow & \color{red}{-77} x& = & -61 \\\Leftrightarrow & x = \frac{-61}{-77} & & \\\Leftrightarrow & x = \frac{61}{77} & & \\ & V = \left\{ \frac{61}{77} \right\} & \\\end{align}\)
12. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (5x+\frac{3}{4})& = & -6x+\frac{3}{2} \\\Leftrightarrow & 25x+\frac{15}{4}& = & -6x+\frac{3}{2} \\ & & & \text{kgv van noemers 4 en 2 is 4} \\\Leftrightarrow & \color{blue}{4} .(\frac{100}{ \color{blue}{4} }x+ \frac{15}{ \color{blue}{4} })& = & (\frac{-24}{ \color{blue}{4} }x+ \frac{6}{ \color{blue}{4} }). \color{blue}{4} \\\Leftrightarrow & 100x \color{red}{+15} & = & \color{red}{-24x} +6 \\\Leftrightarrow & 100x \color{red}{+15} \color{blue}{-15} \color{blue}{+24x} & = & \color{red}{-24x} +6 \color{blue}{+24x} \color{blue}{-15} \\\Leftrightarrow & 100x+24x& = & 6-15 \\\Leftrightarrow & \color{red}{124} x& = & -9 \\\Leftrightarrow & x = \frac{-9}{124} & & \\ & V = \left\{ \frac{-9}{124} \right\} & \\\end{align}\)