The word ""equation"" says that a certain equality registers. In it there are known and unknown sizes. There are equations of different type - logarithmic, indicative, trigonometrical and others. Let's consider how to learn to solve the equations, on the example of the linear equations.
Instruction
1. Learn to solve the simplest linear equation of a type of ax+b=0. x is unknown which should be found. The equations in which x can be only in the first degree, any squares and cubes are called linear. an and b - any numbers, and a cannot equal 0. If an or b are presented in the form of fractions, then in a fraction denominator never happens x. Otherwise not the linear equation can turn out. The linear equation simply is solved. We transfer b to other partysignequality . At the same time the sign which faced b changes on opposite. There was plus - there will be minus. We receive ax=-b. Now we find x for what we divide both parts of equality into a. We receive x=-b/a.
2. To solve more difficult equations, remember the 1st identical transformation. Its sense in the following. To both members of equation it is possible to add the same number or expression. And by analogy - from both members of equation it is possible to take away the same number or expression. Let there is an equation 5x+4=8. Let's take away from the left and right part the same expression (5x+4). We receive the 5x+4th (5x+4) =8-(5x+4). After removal of brackets has 5x+4-5x-4=8-5x-4. As a result it turns out 0=4-5x. At the same time the equation in a different way looks, but its essence remained the same. The initial and final equations are called identically equal.
3. Remember the 2nd identical transformation. Both members of equation can be increased by the same number or expression. By analogy - both members of equation can be divided into the same number or expression. Naturally, it is not necessary to multiply or divide into 0. Let there is an equation 1=8 / (5x+4). Let's increase both parts by the same expression (5x+4). We receive 1 * (5x+4)= (8 * (5x+4)) / (5x+4). After reduction we receive 5x+4=8.
4. Learn to lead by means of simplifications and transformations the linear equations to a familiar look. Let there is an equation (2x+4)/3-(5x-2) of/2=11+ (x-4)/6. This equation precisely is linear because x is in the first degree and in denominators of fractions x is absent. But the equation is not similar to the elementary, sorted on the 1st step. Let's apply the 2nd identical transformation. Let's increase both members of equation by number 6 - a common denominator of all fractions. We receive 6 * (2x+4)/3-6 * (5x-2)/2=6*11+6 * (x-4)/6. After reduction of numerator and a denominator we have 2 * (2x+4)-3 * (5x-2) =66+1 * (x-4). Let's remove the brackets 4x+8-15x+6=66+x-4. As a result 14-11x=62+x. Let's apply the 1st identical transformation. Let's take away from the left and right part expression (62+x). We receive 14-11x-(62+x) =62+x-(62+x). As a result 14-11x-62-x=0. We receive - 12x-48=0. And it is the simplest linear equation which solution is sorted on the 1st step. We presented difficult initial expression with fractions in the usual form, using identical transformations.