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Linear(Simple) Equations
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Linear(Simple) Equations: Problems with Solutions
Problem 1
Find the solution [tex]n[/tex] to the equation [tex]n + 2 = 6[/tex]
Solution:
By substracting 2 from both sides, we get
[tex]n + 2 - 2 = 6 - 2[/tex],
or [tex]n = 4[/tex]
Problem 2
Solve the equation [tex]z - 5 = 6[/tex].
Solution:
By adding [tex]5[/tex] to both sides, we get [tex]z=5+6[/tex], or [tex]z=11[/tex].
Problem 3
Solve the equation [tex]5 - t = 0[/tex].
Solution:
We add [tex]t[/tex] to both sides of the equation, and we get [tex]5-t+t=0+t[/tex], or [tex]t=5[/tex].
Problem 4
Solve the equation [tex]n+7=13[/tex].
Solution:
Subtract [tex]7[/tex] from both sides:
[tex]n+7-7=13-7 \Rightarrow n=6[/tex]
Problem 5
Solve the equation [tex]x+34=82[/tex].
Solution:
[tex]x=82-34=48[/tex]
Problem 6
Solve the linear equation [tex]3-a=2a[/tex].
Solution:
By adding [tex]a[/tex] to both sides, we get [tex]3-a+a=2a+a[/tex], or [tex]3a=3[/tex]. The we divide both sides by 3 to reach the final answer, [tex]a=1[/tex].
Problem 7
Solve the linear equation [tex]2x-20=10[/tex]
Solution:
[tex]2x=10+20[/tex]
[tex]2x=30[/tex]
[tex]x=15[/tex]
Problem 8
Solve the equation [tex]20 - 2x = 18[/tex]
Solution:
[tex]20 - 2x = 18[/tex]
[tex]20 - 18 = 2x[/tex]
[tex]2=2x[/tex]
[tex]x=1[/tex]
Problem 9
Find [tex]c[/tex], if [tex]5c - 2 = 33[/tex].
Solution:
We add 2 to both sides and get [tex]5c-2+2=33+2[/tex], or [tex]5c=35[/tex]. We divide both sides by 5 in order to get [tex]c=7[/tex].
Problem 10
Solve the equation [tex]4x=72[/tex]
Solution:
We divide both sides of the equation with 4:
[tex]\frac{4x}{4}=\frac{72}{4}[/tex]
[tex]x=18[/tex]
Problem 11
Solve the equation [tex]13x-15=24[/tex]
Solution:
[tex]13x=24+15[/tex]
[tex]13x=39[/tex]
[tex]x=3[/tex]
Problem 12
Find the solution b to the equation [tex]\frac{b}{3}=3[/tex].
Solution:
We multiply both sides of the equation by [tex]3[/tex] in order to get [tex]\frac{b}{3}\cdot3=3\cdot3[/tex], or [tex]b=9[/tex].
Problem 13
Solve the linear equation [tex]2x+2=40[/tex]
Solution:
[tex]2x=40-2[/tex]
[tex]2x=38[/tex]
[tex]x=19[/tex]
Problem 14
Solve the equation [tex]3x + 1 = 16[/tex].
Solution:
By substracting 1 from both sides, we get [tex]3x+1-1=16-1[/tex], or [tex]3x=15[/tex]. We divide both sides by 3 in order to get the value for x: [tex]x=5[/tex].
Problem 15
Solve the equation [tex]2x + 5 = 9[/tex]
Solution:
By substracting 5 from both sides of the equation, we get [tex]2x+5-5=9-5[/tex], or [tex]2x=4[/tex]. Then we divide both sides by 2 to reach the solution, [tex]x=2[/tex].
Problem 16
Solve the equation [tex]\text{ m}+10=3\text{ m}[/tex].
Solution:
By substracting [tex]\text{ m}[/tex] from both sides, we get [tex]\text{ m}+10-\text{ m}=3\text{ m}-\text{ m}[/tex], so [tex]10=2\text{ m}[/tex] or [tex]2\text{ m}=10[/tex]. Dividing both sides of the equation by 2, we find that [tex]\text{ m}=5[/tex]
Problem 17
Solve the linear equation [tex]19z=38+6\times 19[/tex]
Solution:
We divide both sides by 19:
[tex]\frac{19z}{19}=\frac{38}{19}+6\cdot\frac{19}{19}[/tex]
[tex]z=2+6[/tex]
[tex]z=8[/tex]
Problem 18
Find the solution [tex]y[/tex] to the linear equation [tex]2y+6=y+11[/tex].
Solution:
First, we substract [tex]6[/tex] from both sides to get [tex]2y+6-6=y+11-6[/tex], or [tex]2y=y+5[/tex].
Then we substract [tex]y[/tex] from both sides to reach the final solution, [tex]y=5[/tex].
Problem 19
Solve the equation [tex]\frac{1}{x}=\frac{1}{5}[/tex]
Solution:
The equation is defined for [tex]x \ne 0[/tex]. We cross multiply and get
[tex]x=5[/tex], which is a valid value for [tex]x[/tex] and the solution.
Problem 20
Find the solution [tex]x[/tex] to the equation [tex]x-1+x-2+x-3=0[/tex]
Solution:
[tex]x+x+x-1-2-3=0[/tex]
[tex]3x-6=0[/tex]
[tex]3x=6[/tex]
[tex]x=2[/tex]
Problem 21
Find the solution [tex]y[/tex] to the equation [tex]y+10=13y-74[/tex]
Solution:
[tex]10+74=13y-y[/tex]
[tex]12y=84[/tex]
[tex]y=7[/tex]
Problem 22
Solve the equation [tex]3c=8c[/tex].
Solution:
We substract [tex]3c[/tex] from both sides of the equation to get [tex]3c-3c=8c-3c[/tex], or [tex]5c=0[/tex]. The we divide both sides by 5 and the answer is [tex]c=0[/tex].
Problem 23
Solve the linear equation [tex]\frac{3}{8}a=3[/tex]
Solution:
First we multiply both sides by 8 to get free of the denominator. That yields [tex]\frac{3}{8}a\cdot8=3\cdot8[/tex], or [tex]3a=24[/tex]. Then we divide both sides by 3 to get [tex]a=8[/tex].
Problem 24
Solve the equation [tex]4x - 9 = 2x + 7[/tex].
Solution:
[tex]4x-9=2x+7[/tex]
[tex]2x=16[/tex]
[tex]x=\frac{16}{2}[/tex]
[tex]x=8[/tex]
Problem 25
If [tex]5x+12=3x-24[/tex], determine the value of [tex]x[/tex].
Solution:
[tex]5x-3x=-24-12[/tex]
[tex]2x=-36[/tex]
[tex]x=-18[/tex]
Problem 26
Find the solution to the equation [tex]\frac{x}{5}=8[/tex].
Solution:
We multiply both sides by 5: [tex]5\cdot\frac{x}{5}=8\cdot5[/tex] in order to get [tex]x=40[/tex].
Problem 27
If [tex]28 = 10y - 3y[/tex], find [tex]y[/tex].
Solution:
We substract on the right side: [tex]28=7y[/tex]. Then we divide by 7 in order to get the solution, [tex]y=4[/tex].
Problem 28
-2x - 6 = 2
Solution:
-2x = 2 + 6
-2x = 8
x= -8 : 2
x= -4
Problem 29 sent by Bhoomi Bembde
If [tex]19 + 19x = 3x - 9[/tex] find the value of [tex]x[/tex].
Solution:
[tex]19 + 19x = 3x - 9[/tex]
[tex]19x - 3x = -9 - 19[/tex]
[tex]16x = -28[/tex]
[tex]x = -28 \div 16 = -\frac{7}{4} = -1.75[/tex]
Problem 30 sent by Oshin Patel
If [tex]x+7=16[/tex], find the value of the expression [tex]8x-72[/tex].
Solution:
[tex]x=16-7=9[/tex]
[tex]8x-72=8\times 9-72=72-72=0[/tex]
Problem 31
Solve the equation [tex]3x+3=7x-9[/tex].
Solution:
[tex]3x+3=7x-9 \iff 3+9=7x-3x \iff 12=4x \iff x=3[/tex]
Problem 32
Factor the polynomial by taking out the common factor: [tex]x(a+3)-y(a+3)[/tex]
[tex](a+3)(x+y)[/tex]
[tex](a+3)(x-y)[/tex]
[tex](a-3)(x-y)[/tex]
[tex](x+3)(a-y)[/tex]
Solution:
The common factor is [tex](a+3)[/tex]:
[tex]x(a+3)-y(a+3)=(a+3)(x-y)[/tex]
Problem 33 sent by Samarth Shivananda Mathad
The sum of two numbers is [tex]45[/tex]. One of the numbers is [tex]9[/tex] greater than the other. Find the numbers.
[tex]19, 28[/tex]
[tex]15, 30[/tex]
[tex]18, 27[/tex]
[tex]17, 28[/tex]
Solution:
Let one number be [tex]x[/tex]; then the other is [tex]x+9[/tex].
[tex]x+(x+9)=45 \Rightarrow 2x=36 \Rightarrow x=36\div 2=18, x+9=27[/tex]
Problem 34
Solve the equation:
[tex](2x+1)^2-(2x-1)^2=-x[/tex]
Solution:
[tex]4x^2+4x+1-(4x^2-4x+1)=-x[/tex]
[tex]8x=-x \Rightarrow 9x=0 \Rightarrow x=0[/tex]
Problem 35
Solve the equation:
[tex](3t-2)(t+2)\left(\frac{3}{2}-t\right)=0[/tex]
[tex]t\in\left\{\frac{2}{3}, -2, \frac{3}{2}\right\}[/tex]
[tex]t\in\left\{-\frac{2}{3}, 2, -\frac{3}{2}\right\}[/tex]
[tex]t\in\left\{\frac{3}{2}, -2, \frac{2}{3}, 0\right\}[/tex]
[tex]t\in\left\{\frac{2}{3}, 2\right\}[/tex]
Solution:
A product is [tex]0[/tex] when at least one of the factors is [tex]0[/tex]:
[tex]3t-2=0 \Rightarrow t=\frac{2}{3}[/tex]
[tex]t+2=0 \Rightarrow t=-2[/tex]
[tex]\frac{3}{2}-t=0 \Rightarrow t=\frac{3}{2}[/tex]
Problem 36
The equation [tex]x^2=x(x+4)[/tex] is equivalent to the equation:
[tex]0\times x=4[/tex]
[tex]5x=0[/tex]
[tex]4x=1[/tex]
[tex]x+4=x+4[/tex]
Solution:
[tex]x^2=x^2+4x \Rightarrow 4x=0 \Rightarrow x=0[/tex]
The equation [tex]5x=0[/tex] also has the single root [tex]x=0[/tex], so the two equations are equivalent.
Problem 37
Of four donors, the second gave twice as much as the first, the third three times as much as the second, and the fourth four times as much as the third. Together they gave [tex]132[/tex] coins. How much did the first donor give? (An ancient Indian problem)
Solution:
If the first gave [tex]x[/tex], the others gave [tex]2x[/tex], [tex]6x[/tex] and [tex]24x[/tex]:
[tex]x+2x+6x+24x=132 \Rightarrow 33x=132 \Rightarrow x=4[/tex]
Problem 38
A bag of sugar is placed on one pan of a balance. On the other pan there are three quarters of the same bag and [tex]12[/tex] ounces. The balance is level. How much does one bag of sugar weigh?
[tex]48[/tex] oz
[tex]36[/tex] oz
[tex]16[/tex] oz
[tex]60[/tex] oz
Solution:
[tex]x=\frac{3}{4}x+12 \Rightarrow \frac{1}{4}x=12 \Rightarrow x=48[/tex] ounces (that is, [tex]3[/tex] pounds).
Problem 39
If we add [tex]12[/tex] to one third of my father's age, we get my age. I am [tex]32[/tex] years old. How old is my father?
Solution:
[tex]\frac{x}{3}+12=32 \Rightarrow \frac{x}{3}=20 \Rightarrow x=60[/tex]
Problem 40
The sum of [tex]7[/tex] consecutive integers is [tex]7[/tex]. What is the smallest of these numbers?
Solution:
Let the numbers be [tex]n, n+1, \ldots, n+6[/tex]:
[tex]7n+(1+2+3+4+5+6)=7 \Rightarrow 7n+21=7 \Rightarrow n=-2[/tex]
The numbers are [tex]-2[/tex], [tex]-1[/tex], [tex]0[/tex], [tex]1[/tex], [tex]2[/tex], [tex]3[/tex], [tex]4[/tex].
Problem 41
I thought of two consecutive even numbers. I multiplied the first by [tex]5[/tex] and divided the second by [tex]4[/tex]. The sum of the results is [tex]74[/tex]. Which numbers did I think of?
[tex]12[/tex] and [tex]14[/tex]
[tex]14[/tex] and [tex]16[/tex]
[tex]16[/tex] and [tex]18[/tex]
[tex]10[/tex] and [tex]12[/tex]
Solution:
[tex]5n+\frac{n+2}{4}=74 \Rightarrow 20n+n+2=296 \Rightarrow 21n=294 \Rightarrow n=14[/tex]
The numbers are [tex]14[/tex] and [tex]16[/tex].
Problem 42
One side of a triangle is [tex]3[/tex] inches shorter than another side and half as long as the third side. Find the sides of the triangle if its perimeter is [tex]23[/tex] inches.
[tex]5, 8, 10[/tex]
[tex]4, 7, 12[/tex]
[tex]6, 9, 8[/tex]
[tex]5, 9, 9[/tex]
Solution:
Let the shortest side be [tex]a[/tex]; the others are [tex]a+3[/tex] and [tex]2a[/tex]:
[tex]a+(a+3)+2a=23 \Rightarrow 4a=20 \Rightarrow a=5[/tex]
The sides are [tex]5[/tex], [tex]8[/tex] and [tex]10[/tex].
Problem 43
Find two positive numbers whose ratio is [tex]4:9[/tex] and whose difference is [tex]165[/tex].
[tex]66[/tex] and [tex]231[/tex]
[tex]120[/tex] and [tex]285[/tex]
[tex]132[/tex] and [tex]297[/tex]
[tex]44[/tex] and [tex]99[/tex]
Solution:
Let the numbers be [tex]4k[/tex] and [tex]9k[/tex]:
[tex]9k-4k=165 \Rightarrow 5k=165 \Rightarrow k=33[/tex]
The numbers are [tex]4\times 33=132[/tex] and [tex]9\times 33=297[/tex].
Problem 44
If [tex]\frac{3a-2b}{2}=\frac{5}{4}[/tex], find [tex]12a-8b[/tex].
Solution:
[tex]3a-2b=\frac{5}{2}[/tex]
[tex]12a-8b=4(3a-2b)=4\times \frac{5}{2}=10[/tex]
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