Linear(Simple) Equations: Difficult Problems with Solutions

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

Solution:
[tex]2(x+50)=120 \Rightarrow 2x+100=120 \Rightarrow 2x=20 \Rightarrow x=10[/tex]

Solution:
We first remove the parentheses on the left side and get [tex]3n+3⋅15=30+6n[/tex], or [tex]3n+45=30+6n[/tex]. We substract [tex]3n[/tex] from both sides and get [tex]45=30+3n[/tex]. We substract 30 from both sides and get [tex]15=3n[/tex]. By dividing both sides by 3, we get the final answer [tex]n=5[/tex].
Second solution: we divide both sides by 3, getting [tex]n+15=10+2n[/tex]. Substracting [tex]n[/tex] from both sides, we get [tex]15=10+n[/tex]. 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 [tex]x+3⋅10=3⋅2x[/tex], or [tex]x+30=6x[/tex]. By substracting x from both sides we get [tex]30=5x[/tex]. Dividing both sides by 5 gives us the answer, [tex]x=6[/tex].

Solution:
We multiply both sides by 5 to get [tex]3x-2⋅5=2x[/tex]. We substract [tex]2x[/tex] from both sides and the equation becomes [tex]x-10=0[/tex], which means [tex]x=10[/tex] 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 [tex]3x+2x=60[/tex]. We have [tex]5x=60[/tex], we divide both sides by 5 and we get [tex]x=12[/tex]

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 - 4x + 18 = 36 + 4x + 18 - 6x
-4x - 4x + 6x = 36 + 18 - 12 - 18
-2x = 24
-x = 12
x = -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 [tex]x[/tex] and therefore a solution.

Solution:
Let us denote the number with [tex]x[/tex]. 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 [tex]7x+10x+14x=155[/tex], or [tex]31x=155[/tex], which means that [tex]x=5[/tex].

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 [tex]4x-3x=24[/tex], which means that [tex]x=24[/tex].

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 [tex]2x-4=10[/tex], or [tex]2x=14[/tex]. Division by 2 gives us [tex]x=7[/tex].

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 [tex]2(5y+4)=36+3(2y+4)[/tex]. Removing the parentheses, we get [tex]10y+8=36+6y+12[/tex]. By substracting [tex]6y[/tex] from both sides, we get [tex]4y+8=48[/tex], or [tex]4y=40[/tex]. Dividing by four gives us [tex]y=10[/tex].

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 [tex]x[/tex].

Solution:
[tex]x^2-4=(x+2)(x-2)=0 \Rightarrow x_1=-2, x_2=2[/tex]
Answer: [tex]-2[/tex]

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

Solution:
Multiply both sides by [tex]6[/tex]:
[tex]3(x-4)-2(2x+5)=6-(x+4)[/tex]
[tex]3x-12-4x-10=6-x-4 \Rightarrow -x-22=2-x \Rightarrow -22=2[/tex]
This is false for every [tex]x[/tex], so the equation has no solution.

Solution:
Let the second shelf have [tex]x[/tex] books:
[tex](x-10)+x+(x+5)=25 \Rightarrow 3x-5=25 \Rightarrow x=10[/tex]
Then the first shelf would have [tex]10-10=0[/tex] books – it would be empty. So it is not possible.

Solution:
[tex]\frac{x}{6}+\frac{x}{12}+\frac{x}{7}+5+\frac{x}{2}+4=x[/tex]
Multiply by [tex]84[/tex] (the least common multiple of [tex]6[/tex], [tex]12[/tex], [tex]7[/tex] and [tex]2[/tex]):
[tex]14x+7x+12x+420+42x+336=84x \Rightarrow 75x+756=84x \Rightarrow 9x=756 \Rightarrow x=84[/tex]

Solution:
[tex]x+x+\frac{x}{2}+\frac{x}{4}+1=100[/tex]
[tex]\frac{11x}{4}=99 \Rightarrow x=36[/tex]

Solution:
After the first step [tex]x-3[/tex] apples remain; after taking a third of them, [tex]\frac{2}{3}(x-3)[/tex] remain:
[tex]\frac{2}{3}(x-3)-3=\frac{x}{2}[/tex]
[tex]4(x-3)-18=3x \Rightarrow 4x-30=3x \Rightarrow x=30[/tex]
Check: [tex]30-3=27[/tex], [tex]27-9=18[/tex], [tex]18-3=15=\frac{30}{2}[/tex]

Solution:
After spending a quarter, [tex]\frac{3}{4}x[/tex] remained; after losing a third of that, [tex]\frac{2}{3}\times \frac{3}{4}x=\frac{x}{2}[/tex] remained:
[tex]\frac{x}{2}=300 \Rightarrow x=600[/tex]

Solution:
Let the units digit be [tex]b[/tex]; then the number is [tex]10(b+2)+b=11b+20[/tex]:
[tex]7\big((b+2)+b\big)=11b+20 \Rightarrow 14b+14=11b+20 \Rightarrow b=2[/tex]
The number is [tex]42[/tex] (check: [tex]4+2=6=42\div 7[/tex]).

Solution:
A table costs [tex]0. 92\times 25=23[/tex] dollars. Let [tex]t[/tex] be the number of tables:
[tex]23t+25(72-t)=1720 \Rightarrow 1800-2t=1720 \Rightarrow t=40[/tex]

Solution:
Let the July profit be [tex]x[/tex]; then June: [tex]0. 85x[/tex], August: [tex]1. 4x[/tex]:
[tex]0. 85x+x+1. 4x=11\,375 \Rightarrow 3. 25x=11\,375 \Rightarrow x=3500[/tex]
June: [tex]0. 85\times 3500=2975[/tex], August: [tex]1. 4\times 3500=4900[/tex].

Solution:
[tex](n+2)^2-n^2=32 \Rightarrow 4n+4=32 \Rightarrow n=7[/tex]
The numbers are [tex]7[/tex] and [tex]9[/tex].

Solution:
[tex](2x-1)^3=8x^3-12x^2+6x-1[/tex]
[tex]2x(2x+3)(2x-3)=2x(4x^2-9)=8x^3-18x[/tex]
[tex](6x-5)(2x+1)=12x^2-4x-5[/tex]
[tex]-12x^2+24x-1=19-12x^2+4x[/tex]
[tex]20x=20 \Rightarrow x=1[/tex]

Solution:
[tex](1)\ 3x-4=3x-17 \Rightarrow -4=-17[/tex] – III
[tex](2)\ -7x-22=x+2 \Rightarrow -8x=24 \Rightarrow x=-3[/tex] – I
[tex](3)\ x^2-2x-x^2+9=2x+9 \Rightarrow -4x=0 \Rightarrow x=0[/tex] – IV
[tex](4)\ 4-12x+9x^2-3x=9x^2-15x+4 \Rightarrow 0=0[/tex] – II

Solution:
We need the [tex]x[/tex] for which [tex]5x-4=6[/tex]:
[tex]5x=10 \Rightarrow x=2[/tex]
[tex]f(6)=2\times 2+1=5[/tex]