Learn Linear Equation in one variable with Free Lessons & Tips

To solve the equation 3x−4=123x−4=12, you want to isolate the variable xx. Follow these steps:

1. Add 4 to both sides of the equation: 3x−4+4=12+43x−4+4=12+4 3x=163x=16

2. Divide both sides by 3 to solve for xx: 3x3=16333x=316 x=163x=316

So, the solution to the equation 3x−4=123x−4=12 is x=163x=316.

To solve the equation 5x−9=85x−9=8, you can follow these steps:

1. Add 9 to both sides of the equation: 5x−9+9=8+95x−9+9=8+9 5x=175x=17

2. Divide both sides by 5 to solve for xx: 5x5=17555x=517 x=175x=517

So, the solution to the equation 5x−9=85x−9=8 is x=175x=517.

Let's denote the rational number as aa, which is −83−38. We are looking for the value to subtract from thrice aa to get 5225.

The expression for thrice aa is 3a3a, and we want to find bb such that:

3a−b=523a−b=25

Now, substitute a=−83a=−38 into the equation:

3(−83)−b=523(−38)−b=25

Simplify the expression:

−8−b=52−8−b=25

Now, isolate bb by adding 8 to both sides of the equation:

−8−b+8=52+8−8−b+8=25+8

−b=212−b=221

Multiply both sides by -1 to solve for bb:

b=−212b=−221

So, subtracting −212−221 from thrice −83−38 will give 5225.

Let's represent the three consecutive multiples of 7 as 7n7n, 7n+77n+7, and 7n+147n+14, where nn is an integer. The sum of these multiples is given as 63:

7n+(7n+7)+(7n+14)=637n+(7n+7)+(7n+14)=63

Combine like terms:

21n+21=6321n+21=63

Subtract 21 from both sides of the equation:

21n=4221n=42

Divide both sides by 21:

n=2n=2

Now that we have the value of nn, we can find the three consecutive multiples of 7:

1. 7n=7×2=147n=7×2=14
2. 7n+7=14+7=217n+7=14+7=21
3. 7n+14=14+14=287n+14=14+14=28

So, the three consecutive multiples of 7 whose sum is 63 are 14, 21, and 28.

Let's denote the present age of Sita as SS and the present age of Sita's father as FF.

According to the given information:

1. The present age of Sita's father (FF) is three times the present age of Sita (SS): F=3SF=3S

2. After six years, the sum of their ages will be 69 years:

Now, substitute the expression for FF from the first equation into the second equation:

Combine like terms:

Subtract 12 from both sides:

Divide by 4:

S=574S=457

Now, we know the present age of Sita (SS). To find the present age of Sita's father (FF), substitute SS into the equation F=3SF=3S:

F=3(574)F=3(457)

F=1714F=4171

So, the present age of Sita is 574457 years, and the present age of Sita's father is years.

Let the original two-digit number be 10a+b10a+b, where aa is the digit in the tens place and bb is the digit in the units place.

According to the given information:

1. The digits of the original number differ by 3, so a−b=3a−b=3.

2. If the digits are interchanged, we get the number 10b+a10b+a.

3. The resulting number, when added to the original number, is 121: (10a+b)+(10b+a)=121(10a+b)+(10b+a)=121

Now, we can set up and solve these equations.

Equation 1: a−b=3a−b=3

Equation 2: (10a+b)+(10b+a)=121(10a+b)+(10b+a)=121

Combine like terms in Equation 2: 11a+11b=12111a+11b=121

Divide both sides by 11: a+b=11a+b=11

Now, we have a system of two equations:

a−b=3a−b=3 a+b=11a+b=11

Adding the two equations to eliminate bb: (a−b)+(a+b)=3+11(a−b)+(a+b)=3+11 2a=142a=14

Divide by 2: a=7a=7

Now that we know aa, we can substitute it back into Equation 1 to find bb: 7−b=37−b=3 b=4b=4

So, the original two-digit number is 10a+b=10×7+4=70+4=7410a+b=10×7+4=70+4=74.

To solve the equation x−2x+1=12x+1x−2=21, you can follow these steps:

1. Cross-multiply to eliminate the fractions: (x−2)×2=1×(x+1)(x−2)×2=1×(x+1)

2. Simplify both sides of the equation: 2x−4=x+12x−4=x+1

3. Move all terms involving xx to one side of the equation and constants to the other side: 2x−x=1+42x−x=1+4

4. Simplify: x=5x=5

So, the solution to the equation x−2x+1=12x+1x−2=21 is x=5x=5.

Let's denote Sanjay's present age as SS.

According to the given information:

1. After 18 years, Sanjay's age will be S+18S+18.
2. Four years ago, Sanjay's age was S−4S−4.

The problem states that after 18 years, Sanjay will be 3 times as old as he was 4 years ago. So, we can set up the equation:

S+18=3×(S−4)S+18=3×(S−4)

Now, solve for SS:

S+18=3S−12S+18=3S−12

Subtract SS from both sides:

18=2S−1218=2S−12

Add 12 to both sides:

30=2S30=2S

Divide by 2:

S=15S=15

So, Sanjay's present age is 15 years.

Let's denote the two numbers as xx and yy, where xx is one number, yy is the other number, and x>yx>y.

According to the given information:

1. The sum of the two numbers is 30: x+y=30x+y=30

2. The ratio of the two numbers is 2/32/3: xy=23yx=32

Now, we can set up a system of equations and solve for xx and yy:

Equation 1: x+y=30x+y=30

Equation 2: xy=23yx=32

To solve for xx and yy, you can use substitution or elimination. Let's use substitution:

From Equation 1, we can express xx in terms of yy: x=30−yx=30−y

Now substitute this expression for xx into Equation 2: 30−yy=23y30−y=32

Cross-multiply to eliminate fractions: 3(30−y)=2y3(30−y)=2y

Simplify and solve for yy: 90−3y=2y90−3y=2y 90=5y90=5y y=18y=18

Now that we have yy, substitute it back into Equation 1 to find xx: x+18=30x+18=30 x=12x=12

So, the two numbers are 12 and 18.

Let's denote the denominator of the fraction as DD. According to the given information:

1. The numerator is 2 less than the denominator, so the numerator is D−2D−2.
2. If one is added to the denominator, the fraction becomes 1221.

So, we can set up the equation:

D−2D+1=12D+1D−2=21

Now, cross-multiply to eliminate fractions:

2(D−2)=D+12(D−2)=D+1

Expand and simplify:

2D−4=D+12D−4=D+1

Subtract DD from both sides:

D−4=1D−4=1

Add 4 to both sides:

D=5D=5

Now that we have the value for the denominator (DD), we can find the numerator:

N=D−2N=D−2 N=5−2N=5−2 N=3N=3

So, the fraction is 3553.