Algebra
1.1. Extra Oefenopgaven
Geef bij onderstaande waarden steeds de juiste waarde van x.
| \[x-6 = 8-x\] | \(x =\) |
- \[x - 6 = 8 - x\]
- \[ +x↓+x \]
- \[2x-6=8\]
- \[ +6↓+6 \]
- \[2x=14\]
- \[ :2↓:2 \]
- \[x=7\]
| \[(x+3)2=28\] | \(x =\) |
- \[(x+3)2=28\]
- \[ :2↓:2 \]
- \[x+3=14\]
- \[ -3↓-3 \]
- \[x=11\]
| \[ x+1 = \frac{4x}{3} \] | \(x =\) |
- \[ x+1 = \frac{4x}{3} \]
- \[ \cdot 3↓\cdot 3 \]
- \[ 3x+3 = 4x\]
- \[ -3x↓-3x \]
- \[ x=3 \]
| \[ 4(x+3)=60 \] | \(x =\) |
- \[ 4(x+3)=60 \]
- \[ :4↓:4 \]
- \[ x+3=15 \]
- \[ -3↓-3 \]
- \[ x=12 \]
| \[ 2(3x+2)=34 \] | \(x =\) |
- \[ 2(3x+2)=34 \]
- \[ :2↓:2 \]
- \[ 3x+2=17 \]
- \[ -2↓-2 \]
- \[ 3x=15 \]
- \[ :3↓:3 \]
- \[ x=5 \]
| \[ \frac{6(x+1)}{3} = 10 \] | \(x =\) |
- \[ \frac{6(x+1)}{3} = 10 \]
- \[ \cdot 3↓\cdot 3 \]
- \[ 6(x+1)=30 \]
- \[ :6↓:6 \]
- \[ x+1=5 \]
- \[ -1↓-1 \]
- \[ x=4 \]
| \[ 3(3x+13) = 6(10+x) \] | \(x =\) |
- \[3(3x+13) = 6(10+x)\]
- \[ :3↓:3 \]
- \[3x+13=2(10+x)=20+2x\]
- \[ -2x↓-2x \]
- \[x+13=20\]
- \[ -13↓-13 \]
- \[x=7\]
| \[ (x-3)2 = 4 \] | \(x =\) |
- \[(x-3)2 = 4\]
- \[ :2↓:2 \]
- \[x-3=2\]
- \[ +3↓+3 \]
- \[x=5\]
| \[ 2x = 27-x \] | \(x =\) |
- \[2x = 27-x\]
- \[ +x↓+x \]
- \[3x=27\]
- \[ :3↓:3 \]
- \[x=9\]
| \[ \frac{x}{4} + x = 4+x \] | \(x =\) |
- \[ \frac{x}{4} + x = 4+x \]
- \[ -x↓-x \]
- \[ \frac{x}{4} = 4 \]
- \[ \cdot 4↓ \cdot 4 \]
- \[ x=16 \]
| \[ x+1 = \frac{3x+3}{x} \] | \(x =\) |
- \[ x+1 = \frac{3x+3}{x} = \frac{3(x+1)}{x} \]
- \[ :(x+1)↓:(x+1) \]
- \[ \frac{3}{x}=1 \]
- \[ :3↓:3 \]
- \[ \frac{1}{x} = \frac{1}{3} \]
- \[ 1:breuk (breuk omdraaien)↓1:breuk (breuk omdraaien) \]
- \[ x=3 \]
Extra oefenopgaven
Zie je geen vragen? Dan zijn er geen vragen beschikbaar voor het huidig niveau.
| \[ 3x+5=54-4x \] | \(x =\) |
- \[ 3x+5=54-4x \]
- \[ -5↓-5 \]
- \[ 3x=49-4x \]
- \[ +4x↓+4x \]
- \[ 7x=49 \]
- \[ :7↓:7 \]
- \[ x=7 \]
| \[ \frac{2(x+3)}{x}= \frac{6}{x} \] | \(x =\) |
- \[\frac{2(x+3)}{x}= \frac{6}{x}\]
- \[ \cdot x↓\cdot x \]
- \[ 2(x+3)=6\]
- \[ :2↓:2 \]
- \[x+3=3\]
- \[ -3↓-3 \]
- \[x=0\]
| \[ 2x+3=10-5x \] | \(x =\) |
- \[2x+3=10-5x\]
- \[ -3↓-3 \]
- \[2x=7-5x\]
- \[ +5x↓+5x \]
- \[7x=7\]
- \[ :7↓:7 \]
- \[x=1\]
| \[ \frac{(x+3)2}{x-5} = \frac{18}{x-5} \] | \(x =\) |
- \[\frac{(x+3)2}{x-5} = \frac{18}{x-5}\]
- \[ cdot (x-5)↓cdot (x-5) \]
- \[2(x+3)=18\]
- \[ :2↓:2 \]
- \[x+3=9\]
- \[ -3↓-3 \]
- \[x=6\]
| \[ 1 - \frac{1}{x-1} = \frac{1}{x-1} \] | \(x =\) |
- \[1 - \frac{1}{x-1} = \frac{1}{x-1}\]
- \[ +\frac{1}{1-x}↓+\frac{1}{1-x} \]
- \[\frac{2}{x-1}=1\]
- \[ \cdot (x-1)↓\cdot (x-1) \]
- \[x-1=2\]
- \[ +1↓+1 \]
- \[x=3\]
| \[ 8( \frac{x}{5}+1) = 24 \] | \(x =\) |
- \[8( \frac{x}{5}+1) = 24\]
- \[ :8↓:8 \]
- \[\frac{x}{5}+1 = 3\]
- \[ -1↓-1 \]
- \[\frac{x}{5}=2\]
- \[ \cdot 5↓\cdot 5 \]
- \[x=10\]
| \[ \frac{4(x+1)}{3} = \frac{7}{3} + x \] | \(x =\) |
- \[\frac{4(x+1)}{3} = \frac{7}{3} + x\]
- \[ \cdot 3↓\cdot 3 \]
- \[4(x+1)=4x+4=7+3x\]
- \[ -3x↓-3x \]
- \[x+4=7\]
- \[ -4↓-4 \]
- \[x=3\]
| \[ \frac{1+2x}{3x-1} = 1\] | \(x =\) |
- \[\frac{1+2x}{3x-1} = 1\]
- \[ \cdot (3x-1)↓\cdot (3x-1) \]
- \[1+2x=3x-1\]
- \[ +1↓+1 \]
- \[2+2x=3x\]
- \[ -2x↓-2x \]
- \[x=2\]
| \[ 2x = \frac{75+3x}{4} \] | \(x =\) |
- \[2x = \frac{75+3x}{4}\]
- \[ \cdot 4↓\cdot 4 \]
- \[8x = 75+3x\]
- \[ -3x↓-3x \]
- \[5x=75\]
- \[ :5↓:5 \]
- \[x=15\]