First-degree equations with one unknown

Solve the following equation:

$12\left(2X-3\right)=-4\left(3-4X\right)$

Solution:

To solve the equation, we first make the products of the two sides of the equation:

$24X-36=-12+16X$

Next we will group the like terms, so that on the left side of the equation all the unknowns appear, while on the right side of the equation the numbers appear. Remember, when transposing the terms from one side of the equation to the other, their sign will change. That is, if it is adding, it will go to the other side subtracting, and vice versa.

$24X-16X=-12+36$

Then we reduce the like terms:

$8X=24$

Now, to find the value of the unknown, we divide both sides of the equation by 8 and get:

$8X/8=24/8$

$X=3$

Thus, $X=3$ is the solution of the equation.

Answer:

$X=3$

Solve the following equation:

$8\left(2-5X\right)-12\left(1-X\right)=0$

To solve this equation, we first do the product of the left side of the equation, obtaining:

$16-40X-12+12X=0$

Next we group the like terms, so that on the left side of the equation all the unknowns appear, while on the right side of the equation the numbers will appear. Remember, when transposing the terms from one side of the equation to the other, the sign of the terms will change.

$-40X+12X=12-16$

The next step is to reduce the like terms:

$-28X=-4$

Now, to find the value of the unknown, we divide the two sides of the equation by (-28) and we will get:

$-28X/-28=-4/-28$

And finally we reduce the fraction:

$X=\frac{4}{28}=\frac{1}{7}$

Answer:

$X=\frac{1}{7}$

Solve the following equation:

$-6\left(-X-1\right)+10\left(2-X\right)=16$

To solve the equation, we first make the products of the two sides of the equation:

$6X+6+20-10X=16$

We then group the like terms together, so that on the left side of the equation all the unknowns appear, while on the right side of the equation the numbers appear. Remember, when transposing the terms from one side of the equation to the other, their sign will change. That is, if it is adding, it will go to the other side subtracting, and vice versa.

$6X-10X=16-6-20$

The next step is to reduce the like terms:

$-4X=-10$

Now, to find the value of the unknown, we divide both sides of the equation by (-4), and we will get:

$-4X/-4=-\frac{10}{-4}$

$X=\frac{10}{4}=2.5$

Answer:

$X=2.5$

Solve the following equation:

$3\frac{1}{2}\cdot y=21$

Solution

Note that:

$3\frac{1}{2}=\frac{7}{2}$

Thus the equation is equivalent to:

$\frac{7}{2}\cdot y=21$

Now, we divide by 7/2 both sides of the equation and get:

$y=\frac{21}{\frac{7}{2}}=6$

$y=6$

Answer

$y=6$

Solve the following equation:

$4\frac{1}{3}\cdot x=21\frac{2}{3}$

Solution

Note that:

$4\frac{1}{3}=\frac{13}{3}$

y

$21\frac{2}{3}=\frac{65}{3}$

Thus, the equation is equivalent to:

$\frac{13}{3}\cdot x=\frac{65}{3}$

Divide both sides of the equation by:

$\frac{13}{3}$

to simplify

$x=\frac{\frac{65}{3}}{\frac{13}{3}}$

$x=5$

Answer

$x=5$

Solve the following equation:

$3x+4+x+1=9$

Solution

Next we group the like terms, so that on the left side of the equation all the unknowns appear, while on the right side of the equation the numbers appear.

$3x+x=9-4-1$

We add the like terms:

$4x=4$

We divide both sides of the equation by $4$

$4x/4=4/4$

Answer

$x=1$

Solve the following problem:

What is the domain of application of the equation?

$\frac{xyz}{2(3+y)+4}=8$

Solution

We must calculate when the denominator on the right hand side of the equation equals zero, i.e:

$2\left(3+y\right)+4=0$

We multiply by $2$ in the two elements of the parentheses.

$6+2y+4=0$

We add accordingly

$10+2y=0$

We go to $10$ to the right hand section

$2y=-10$

Divide by $2$

$y=-5$

$y\ne-5$

If $Y$ is equal to minus $5$ then the denominator is equal to $0$ and the exercise has no solution.

Answer

$y\ne-5$

It is a mathematical expression consisting of an unknown or variable and numbers in which the value of the variable must be found, which is generally denoted by $X$.

Examples

a) $3x-5=2x+4$.

b) $4-x=10$.

c) $4(x-4)+2=2x$.

Isolating the unknown, that is, leaving it alone somewhere in the equality.

It is a mathematical expression consisting of two unknowns or variables and numbers in which the value of the variables must be found, which are generally denoted by $X$ and $Y$.

Isolating the variable or unknown using operations such as addition, subtraction, multiplication and division.