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Practice
Linear(Simple) Equations
Easy
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Difficult
Linear(Simple) Equations: Difficult Problems with Solutions
Problem 1
Solve the equation
8+2(x-4)=16
.
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
.
Problem 2
Solve the equation
3(n+15)=30+6n
.
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.
Problem 3
Solve the equation: [tex]\frac{x}{3}+10=2x[/tex].
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
.
Problem 4
Solve the linear equation [tex]\frac{3x}{5}-2=\frac{2x}{5}[/tex].
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.
Problem 5
Solve the equation [tex]\frac{x}{2}+\frac{x}{3}=10[/tex].
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
Problem 6
Solve the equation [tex]\frac{x+11}{x+6}=6[/tex].
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.
Problem 7
If [tex]x+1+x+2+x+3+x-1+x-2+x-3=24[/tex], find
x
.
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]
Problem 8
Find the solution
x
to the equation
[tex]x+2x+3x+4x=23780[/tex]
Solution:
[tex]10x=23780[/tex]
[tex]x=2378[/tex]
Problem 9 sent by Dana Kandybaeva
12 – (4х – 18) = (36 + 4х) + (18 – 6х)
Solution:
12 - 4х + 18 = 36 + 4х + 18 - 6х
-4х - 4х + 6х = 36 + 18 - 12 - 18
-2х = 24
-х = 12
х = -12
Problem 10
A number equals 5 times itself. What is the number?
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]
Problem 11 sent by Arnav Tyagi
2(3x - 7) + 4(3x + 2) = 6(5x + 9)
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
Problem 12
Solve the equation [tex]\frac{7x}{15}+\frac{8x}{15}=723[/tex]
Solution:
[tex]\frac{7x+8x}{15}=723[/tex]
[tex]\frac{15x}{15}=723[/tex]
[tex]x=723[/tex]
Problem 13
Solve the equation [tex]\frac{5x+1}{17}=3[/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]
Problem 14
Solve the linear equation [tex]\frac{x+1}{5}+\frac{x+2}{5}=17[/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]
Problem 15
Solve the equation [tex]\frac{x}{10}+\frac{x}{7}+\frac{x}{5}=\frac{31}{14}[/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
.
Problem 16
Find the solution
x
to the equation [tex]\frac{x}{3}-\frac{x}{4}=2[/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
4x-3x=24
, which means that
x=24
.
Problem 17
Find
x
if [tex]\frac{x+5}{2}+\frac{x+8}{4}=6[/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]
Problem 18
Solve the equation [tex]\frac{1}{7}x-\frac{4}{14}=\frac{5}{7}[/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
.
Problem 19
Find the solution
y
to the equation [tex]\frac{5y+4}{9}=2+\frac{2y+4}{6}[/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
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
.
Problem 20
Solve the equation [tex]\frac{2x+1}{x+1}=3[/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
x
.
Problem 21 sent by Prashant Trivedi
Find the smallest root of the equation: [tex]x^2 - 4 = 0[/tex]
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.
Problem 22
A number, multiplied by 5, equals itself minus 48. Which is the number?
Solution:
[tex]5x=x-48[/tex]
[tex]4x=-48[/tex]
[tex]x=-12[/tex]
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