Linear(Simple) Equations: Difficult Problems with Solutions

Solution:
First, we remove the parentheses and get 8+2x-2⋅4=16, or 8+2x-8=16, which gives us 2x=16. We divide by 2 in order to get x=8.

Solution:
The equation is $2(x+50)=120$

$2x+100=120$   Reveal brackets.
$2x=120-100$   Subract 100 from both sides
$2x=20$   Divide both sides by 2.
$x=10$

Solution:
We first remove the parentheses on the left side and get 3n+3⋅15=30+6n, or 3n+45=30+6n. We substract 3n from both sides and get 45=30+3n. We substract 30 from both sides and get 15=3n. By dividing both sides by 3, we get the final answer n=5.
Second solution: we divide both sides by 3, getting n+15=10+2n. Substracting n from both sides, we get 15=10+n. The final answer is reached by substracting 10 from both sides.

Solution:
We multiply both sides by 3 to get free of the denominator. This gives us x+3⋅10=3⋅2x, or x+30=6x. By substracting x from both sides we get 30=5x. Dividing both sides by 5 gives us the answer, x=6.

Solution:
We multiply both sides by 5 to get 3x-2⋅5=2x. We substract 2x from both sides and the equation becomes x-10=0, which means x=10 is the solution.

Solution:
We must first multiply the equation by the least common multiple of 2 and 3. Since they are relatively prime numbers, it is their product 6. Multiplying the equation by 6 gives us [tex]\frac{x}{2}\cdot6+\frac{x}{3}\cdot6=10\cdot6[/tex], or 3x+2x=60. We have 5x=60, we divide both sides by 5 and we get x=12

Solution:
[tex]x+x+x+x+x+x+1+2+3-1-2-3=24[/tex]
[tex]6x+0=24[/tex]
[tex]x=4[/tex]

Solution:
[tex]10x=23780[/tex]
[tex]x=2378[/tex]

Solution:
12 - 4х + 18 = 36 + 4х + 18 - 6х
-4х - 4х + 6х = 36 + 18 - 12 - 18
-2х = 24
-х = 12
х = -12

Solution:
The equation is defined for [tex]x \ne -6[/tex].
We multiply both sides by [tex]x+6[/tex]:
[tex]\frac{x+11}{x+6}\cdot(x+6)=6(x+6)[/tex]
[tex]x+11=6x+36[/tex]
[tex]6x-x=11-36[/tex]
[tex]x=-5[/tex], which is a valid value for x and therefore a solution.

Solution:
Let us denote the number with x. We have the equation
[tex]x=5x[/tex]
[tex]5x-x=0[/tex]
[tex]4x=0[/tex]
[tex]x=0[/tex]

Solution:
2(3x-7)+4(3x+2)=6(5x+9)
6x - 14 + 12x + 8 = 30x + 54
6x + 12x - 30x = 14 - 8 + 54
-12x = 60
x = 60 ÷ (-12)
x = -5

Solution:
[tex]\frac{7x+8x}{15}=723[/tex]
[tex]\frac{15x}{15}=723[/tex]
[tex]x=723[/tex]

Solution:
We multiply both sides by 17:
[tex]\frac{5x+1}{17}\cdot17=3\cdot17[/tex]
[tex]5x+1=51[/tex]
[tex]5x=50[/tex]
[tex]x=10[/tex]

Solution:
We multiply both sides by 5:
[tex]x+1+x+2=17\cdot 5[/tex]
[tex]2x+3=85[/tex]
[tex]2x=82[/tex]
[tex]x=41[/tex]

Solution:
We first multiply the equation by the least common multiple of the denominators. LCM(14,10,7,5)=70, so we get [tex]70\cdot\frac{x}{10}+70\cdot\frac{x}{7}+70\cdot\frac{x}{5}=70\cdot\frac{31}{14}[/tex], which is equivalent to 7x+10x+14x=155, or 31x=155, which means that x=5.

Solution:
We first find the lowest common multiple of 4 and 3. It is 12. Multiplying both sides by 12 gives us [tex]\frac{x}{3}\cdot12-\frac{x}{4}\cdot12=2\cdot12[/tex], or 4x-3x=24, which means that x=24.

Solution:
The least common multiple of 2 and 4 is obviously 4. We multiply both sides of the equation by 4:
[tex]4\cdot\frac{x+5}{2}+4\cdot\frac{x+8}{4}=4\cdot6[/tex]
[tex]2(x+5)+x+8=24[/tex]
[tex]2x+10+x+8=24[/tex]
[tex]3x+18=24[/tex]. We divide both sides by 3:
[tex]\frac{3x}{3}+\frac{18}{3}=\frac{24}{3}[/tex]
[tex]x+6=8[/tex]
[tex]x=2[/tex]

Solution:
The LCM of the denominators (7 and 14) is 14. We multiply both sides by 14: [tex]14\cdot\frac{1}{7}x-14\cdot\frac{4}{14}=14\cdot\frac{5}{7}[/tex]. We get 2x-4=10, or 2x=14. Division by 2 gives us x=7.

Solution:
First, we find the LCM of the denominators (6 and 9). It is 18. Multiplying both sides by 18 yields [tex]18\cdot\frac{5y+4}{9}=2\cdot18+18\cdot\frac{2y+4}{6}[/tex], which can be also written as 2(5y+4)=36+3(2y+4). Removing the parentheses, we get 10y+8=36+6y+12. By substracting 6y from both sides, we get 4y+8=48, or 4y=40. Dividing by four gives us y=10.

Solution:
The equation is defined for [tex]x \ne -1[/tex]. We multiply both sides by [tex]x+1[/tex]:
[tex]\frac{2x+1}{x+1}\cdot(x+1)=3(x+1)[/tex]
[tex]2x+1=3(x+1)[/tex]
[tex]2x+1=3x+3[/tex]
[tex]x=-2[/tex], which is a valid value for x.

Solution:
[tex]x^2 - 4 = 0[/tex]
(x + 2)(x - 2) = 0
x - 2 = 0 OR x + 2 = 0
x = 2 OR x = -2
x=-2 is smaller so it is the solution of the problem.

Solution:
[tex]5x=x-48[/tex]
[tex]4x=-48[/tex]
[tex]x=-12[/tex]