Vgln. eerste graad (reeks 5)

Hoofdmenu Eentje per keer  🖨

Alles samen. Gebruik stappenplan en ZRM!

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

---

Alles samen. Gebruik stappenplan en ZRM!

Verbetersleutel

1. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (-5x-\frac{4}{3})& = & -7x+\frac{8}{9} \\\Leftrightarrow & -25x-\frac{20}{3}& = & -7x+\frac{8}{9} \\ & & & \text{kgv van noemers 3 en 9 is 9} \\\Leftrightarrow & \color{blue}{9} .(\frac{-225}{ \color{blue}{9} }x- \frac{60}{ \color{blue}{9} })& = & (\frac{-63}{ \color{blue}{9} }x+ \frac{8}{ \color{blue}{9} }). \color{blue}{9} \\\Leftrightarrow & -225x \color{red}{-60} & = & \color{red}{-63x} +8 \\\Leftrightarrow & -225x \color{red}{-60} \color{blue}{+60} \color{blue}{+63x} & = & \color{red}{-63x} +8 \color{blue}{+63x} \color{blue}{+60} \\\Leftrightarrow & -225x+63x& = & 8+60 \\\Leftrightarrow & \color{red}{-162} x& = & 68 \\\Leftrightarrow & x = \frac{68}{-162} & & \\\Leftrightarrow & x = \frac{-34}{81} & & \\ & V = \left\{ \frac{-34}{81} \right\} & \\\end{align}\)
2. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{4} (3x+\frac{4}{5})& = & 5x+\frac{7}{6} \\\Leftrightarrow & 12x+\frac{16}{5}& = & 5x+\frac{7}{6} \\ & & & \text{kgv van noemers 5 en 6 is 30} \\\Leftrightarrow & \color{blue}{30} .(\frac{360}{ \color{blue}{30} }x+ \frac{96}{ \color{blue}{30} })& = & (\frac{150}{ \color{blue}{30} }x+ \frac{35}{ \color{blue}{30} }). \color{blue}{30} \\\Leftrightarrow & 360x \color{red}{+96} & = & \color{red}{150x} +35 \\\Leftrightarrow & 360x \color{red}{+96} \color{blue}{-96} \color{blue}{-150x} & = & \color{red}{150x} +35 \color{blue}{-150x} \color{blue}{-96} \\\Leftrightarrow & 360x-150x& = & 35-96 \\\Leftrightarrow & \color{red}{210} x& = & -61 \\\Leftrightarrow & x = \frac{-61}{210} & & \\ & V = \left\{ \frac{-61}{210} \right\} & \\\end{align}\)
3. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{6} (4x-\frac{3}{5})& = & -5x+\frac{9}{7} \\\Leftrightarrow & 24x-\frac{18}{5}& = & -5x+\frac{9}{7} \\ & & & \text{kgv van noemers 5 en 7 is 35} \\\Leftrightarrow & \color{blue}{35} .(\frac{840}{ \color{blue}{35} }x- \frac{126}{ \color{blue}{35} })& = & (\frac{-175}{ \color{blue}{35} }x+ \frac{45}{ \color{blue}{35} }). \color{blue}{35} \\\Leftrightarrow & 840x \color{red}{-126} & = & \color{red}{-175x} +45 \\\Leftrightarrow & 840x \color{red}{-126} \color{blue}{+126} \color{blue}{+175x} & = & \color{red}{-175x} +45 \color{blue}{+175x} \color{blue}{+126} \\\Leftrightarrow & 840x+175x& = & 45+126 \\\Leftrightarrow & \color{red}{1015} x& = & 171 \\\Leftrightarrow & x = \frac{171}{1015} & & \\ & V = \left\{ \frac{171}{1015} \right\} & \\\end{align}\)
4. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-3} (3x-\frac{4}{5})& = & -5x+\frac{6}{5} \\\Leftrightarrow & -9x+\frac{12}{5}& = & -5x+\frac{6}{5} \\ & & & \text{kgv van noemers 5 en 5 is 5} \\\Leftrightarrow & \color{blue}{5} .(\frac{-45}{ \color{blue}{5} }x+ \frac{12}{ \color{blue}{5} })& = & (\frac{-25}{ \color{blue}{5} }x+ \frac{6}{ \color{blue}{5} }). \color{blue}{5} \\\Leftrightarrow & -45x \color{red}{+12} & = & \color{red}{-25x} +6 \\\Leftrightarrow & -45x \color{red}{+12} \color{blue}{-12} \color{blue}{+25x} & = & \color{red}{-25x} +6 \color{blue}{+25x} \color{blue}{-12} \\\Leftrightarrow & -45x+25x& = & 6-12 \\\Leftrightarrow & \color{red}{-20} x& = & -6 \\\Leftrightarrow & x = \frac{-6}{-20} & & \\\Leftrightarrow & x = \frac{3}{10} & & \\ & V = \left\{ \frac{3}{10} \right\} & \\\end{align}\)
5. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-7} (-3x-\frac{3}{8})& = & -8x+\frac{3}{7} \\\Leftrightarrow & 21x+\frac{21}{8}& = & -8x+\frac{3}{7} \\ & & & \text{kgv van noemers 8 en 7 is 56} \\\Leftrightarrow & \color{blue}{56} .(\frac{1176}{ \color{blue}{56} }x+ \frac{147}{ \color{blue}{56} })& = & (\frac{-448}{ \color{blue}{56} }x+ \frac{24}{ \color{blue}{56} }). \color{blue}{56} \\\Leftrightarrow & 1176x \color{red}{+147} & = & \color{red}{-448x} +24 \\\Leftrightarrow & 1176x \color{red}{+147} \color{blue}{-147} \color{blue}{+448x} & = & \color{red}{-448x} +24 \color{blue}{+448x} \color{blue}{-147} \\\Leftrightarrow & 1176x+448x& = & 24-147 \\\Leftrightarrow & \color{red}{1624} x& = & -123 \\\Leftrightarrow & x = \frac{-123}{1624} & & \\ & V = \left\{ \frac{-123}{1624} \right\} & \\\end{align}\)
6. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-4} (-4x+\frac{4}{7})& = & -3x+\frac{9}{10} \\\Leftrightarrow & 16x-\frac{16}{7}& = & -3x+\frac{9}{10} \\ & & & \text{kgv van noemers 7 en 10 is 70} \\\Leftrightarrow & \color{blue}{70} .(\frac{1120}{ \color{blue}{70} }x- \frac{160}{ \color{blue}{70} })& = & (\frac{-210}{ \color{blue}{70} }x+ \frac{63}{ \color{blue}{70} }). \color{blue}{70} \\\Leftrightarrow & 1120x \color{red}{-160} & = & \color{red}{-210x} +63 \\\Leftrightarrow & 1120x \color{red}{-160} \color{blue}{+160} \color{blue}{+210x} & = & \color{red}{-210x} +63 \color{blue}{+210x} \color{blue}{+160} \\\Leftrightarrow & 1120x+210x& = & 63+160 \\\Leftrightarrow & \color{red}{1330} x& = & 223 \\\Leftrightarrow & x = \frac{223}{1330} & & \\ & V = \left\{ \frac{223}{1330} \right\} & \\\end{align}\)
7. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (-2x-\frac{3}{11})& = & -7x+\frac{8}{11} \\\Leftrightarrow & -10x-\frac{15}{11}& = & -7x+\frac{8}{11} \\ & & & \text{kgv van noemers 11 en 11 is 11} \\\Leftrightarrow & \color{blue}{11} .(\frac{-110}{ \color{blue}{11} }x- \frac{15}{ \color{blue}{11} })& = & (\frac{-77}{ \color{blue}{11} }x+ \frac{8}{ \color{blue}{11} }). \color{blue}{11} \\\Leftrightarrow & -110x \color{red}{-15} & = & \color{red}{-77x} +8 \\\Leftrightarrow & -110x \color{red}{-15} \color{blue}{+15} \color{blue}{+77x} & = & \color{red}{-77x} +8 \color{blue}{+77x} \color{blue}{+15} \\\Leftrightarrow & -110x+77x& = & 8+15 \\\Leftrightarrow & \color{red}{-33} x& = & 23 \\\Leftrightarrow & x = \frac{23}{-33} & & \\\Leftrightarrow & x = \frac{-23}{33} & & \\ & V = \left\{ \frac{-23}{33} \right\} & \\\end{align}\)
8. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-2} (3x+\frac{3}{7})& = & 7x+\frac{8}{3} \\\Leftrightarrow & -6x-\frac{6}{7}& = & 7x+\frac{8}{3} \\ & & & \text{kgv van noemers 7 en 3 is 21} \\\Leftrightarrow & \color{blue}{21} .(\frac{-126}{ \color{blue}{21} }x- \frac{18}{ \color{blue}{21} })& = & (\frac{147}{ \color{blue}{21} }x+ \frac{56}{ \color{blue}{21} }). \color{blue}{21} \\\Leftrightarrow & -126x \color{red}{-18} & = & \color{red}{147x} +56 \\\Leftrightarrow & -126x \color{red}{-18} \color{blue}{+18} \color{blue}{-147x} & = & \color{red}{147x} +56 \color{blue}{-147x} \color{blue}{+18} \\\Leftrightarrow & -126x-147x& = & 56+18 \\\Leftrightarrow & \color{red}{-273} x& = & 74 \\\Leftrightarrow & x = \frac{74}{-273} & & \\\Leftrightarrow & x = \frac{-74}{273} & & \\ & V = \left\{ \frac{-74}{273} \right\} & \\\end{align}\)
9. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-5} (4x+\frac{5}{2})& = & -3x+\frac{4}{5} \\\Leftrightarrow & -20x-\frac{25}{2}& = & -3x+\frac{4}{5} \\ & & & \text{kgv van noemers 2 en 5 is 10} \\\Leftrightarrow & \color{blue}{10} .(\frac{-200}{ \color{blue}{10} }x- \frac{125}{ \color{blue}{10} })& = & (\frac{-30}{ \color{blue}{10} }x+ \frac{8}{ \color{blue}{10} }). \color{blue}{10} \\\Leftrightarrow & -200x \color{red}{-125} & = & \color{red}{-30x} +8 \\\Leftrightarrow & -200x \color{red}{-125} \color{blue}{+125} \color{blue}{+30x} & = & \color{red}{-30x} +8 \color{blue}{+30x} \color{blue}{+125} \\\Leftrightarrow & -200x+30x& = & 8+125 \\\Leftrightarrow & \color{red}{-170} x& = & 133 \\\Leftrightarrow & x = \frac{133}{-170} & & \\\Leftrightarrow & x = \frac{-133}{170} & & \\ & V = \left\{ \frac{-133}{170} \right\} & \\\end{align}\)
10. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-3} (-3x+\frac{5}{4})& = & -7x+\frac{8}{3} \\\Leftrightarrow & 9x-\frac{15}{4}& = & -7x+\frac{8}{3} \\ & & & \text{kgv van noemers 4 en 3 is 12} \\\Leftrightarrow & \color{blue}{12} .(\frac{108}{ \color{blue}{12} }x- \frac{45}{ \color{blue}{12} })& = & (\frac{-84}{ \color{blue}{12} }x+ \frac{32}{ \color{blue}{12} }). \color{blue}{12} \\\Leftrightarrow & 108x \color{red}{-45} & = & \color{red}{-84x} +32 \\\Leftrightarrow & 108x \color{red}{-45} \color{blue}{+45} \color{blue}{+84x} & = & \color{red}{-84x} +32 \color{blue}{+84x} \color{blue}{+45} \\\Leftrightarrow & 108x+84x& = & 32+45 \\\Leftrightarrow & \color{red}{192} x& = & 77 \\\Leftrightarrow & x = \frac{77}{192} & & \\ & V = \left\{ \frac{77}{192} \right\} & \\\end{align}\)
11. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{5} (5x+\frac{2}{3})& = & -9x+\frac{9}{11} \\\Leftrightarrow & 25x+\frac{10}{3}& = & -9x+\frac{9}{11} \\ & & & \text{kgv van noemers 3 en 11 is 33} \\\Leftrightarrow & \color{blue}{33} .(\frac{825}{ \color{blue}{33} }x+ \frac{110}{ \color{blue}{33} })& = & (\frac{-297}{ \color{blue}{33} }x+ \frac{27}{ \color{blue}{33} }). \color{blue}{33} \\\Leftrightarrow & 825x \color{red}{+110} & = & \color{red}{-297x} +27 \\\Leftrightarrow & 825x \color{red}{+110} \color{blue}{-110} \color{blue}{+297x} & = & \color{red}{-297x} +27 \color{blue}{+297x} \color{blue}{-110} \\\Leftrightarrow & 825x+297x& = & 27-110 \\\Leftrightarrow & \color{red}{1122} x& = & -83 \\\Leftrightarrow & x = \frac{-83}{1122} & & \\ & V = \left\{ \frac{-83}{1122} \right\} & \\\end{align}\)
12. \(\text{(1) Haakjes (2) Breuken weg (3) + - (4) . /} \\ \begin{align} & \color{red}{-5} (-4x-\frac{3}{4})& = & 7x+\frac{3}{8} \\\Leftrightarrow & 20x+\frac{15}{4}& = & 7x+\frac{3}{8} \\ & & & \text{kgv van noemers 4 en 8 is 8} \\\Leftrightarrow & \color{blue}{8} .(\frac{160}{ \color{blue}{8} }x+ \frac{30}{ \color{blue}{8} })& = & (\frac{56}{ \color{blue}{8} }x+ \frac{3}{ \color{blue}{8} }). \color{blue}{8} \\\Leftrightarrow & 160x \color{red}{+30} & = & \color{red}{56x} +3 \\\Leftrightarrow & 160x \color{red}{+30} \color{blue}{-30} \color{blue}{-56x} & = & \color{red}{56x} +3 \color{blue}{-56x} \color{blue}{-30} \\\Leftrightarrow & 160x-56x& = & 3-30 \\\Leftrightarrow & \color{red}{104} x& = & -27 \\\Leftrightarrow & x = \frac{-27}{104} & & \\ & V = \left\{ \frac{-27}{104} \right\} & \\\end{align}\)