# Advice 1: How to solve a system of equations for grade 7

The standard system of equations of math for students of the seventh class consists of two equations involving two unknowns. Thus, the task of the student is to find the values of these unknowns, where both equations will become true. This can be done in two main ways. ## The method of substitution

To understand the essence of this method is easiest on the example of solving a typical system includes two equations and requires finding the values of the two unknowns. So, this may be the following system consisting of the equations x + 2y = 6 and x - 3y = -18. In order to solve it by the method of substitution is required in any of the equations to Express one member through another. For example, this can be done using the first equation: x = 6 - 2y.

You must then substitute the resulting expression into the second equation instead of x. The result of this lookup will be the equality of 6 - 2y - 3y = -18. By performing simple arithmetic calculations, this equation is easy to lead to standard view 5y = 24 where y = 4,8. Then the value obtained should be substituted in the expression used for the lookup. Hence x = 6 - 2*4,8 = -3,6.

Then it is advisable to verify the obtained results, substituting them in both equations of the original system. This will give the following equations: -3,6 + 2*4,8 = 6 and -3,6 - 3*4,8 = -18. Both of these equality are true, so you can conclude that the system is solved correctly.

The second method of solving such systems of equations is called the method of addition, which can be illustrated on the basis of the same example. To use, all members of one of the equations is multiplied by a certain factor, with the result that one of them will be the opposite to another. The choice of this ratio is the method of selection, and the same system it is possible to decide using different factors.

In this case, it is advisable to perform a multiplication of the second equation by the coefficient -1. Thus, the first equation retains its original form x + 2y = 6, and the second will take the form -x + 3y = 18. Then you must combine the resulting equations: x + 2y - x + 3y = 6 + 18.

Having made simple calculations, we can obtain an equation of the form 5y = 24, which is similar to the equation resulting from the solution of the system by method of substitution. Accordingly, the roots of such equation will be the same values: x = -3,6, y = 4,8. This clearly demonstrates that both methods are equally applicable for the solution of such systems, and both give the same correct results.

The choice of a particular method may depend on the personal preferences of the student or from a particular expression in which it is easier to Express one through another member or to choose a ratio that will make the members of the two equations are opposite.

# Advice 2 : How to solve example algebra for grade 7

Very often when solving problems on algebra for class 7 the complexity is represented by examples with polynomials. Simplification examples or bringing them to the specified view should know the basic rules of the transformation polynomials. Also the student will need the basics of working with parentheses. Any example can be simplified, reducing the expression by a common divisor, moving the General part of the brackets or by bringing to a common denominator. Upon any conversion of the polynomial is very important to consider the sign of each term. Instruction
1
Note the example given on the sheet. If it is a polynomial highlight in this General part. To do this, find all terms with the same base. The same base members with the same literal part and also with one degree. Such terms are called similar. 2
Fold such members. In this case, consider the signs ahead of them. If before one of them is the " - " sign instead of addition perform subtraction of members and subject to the same sign, write down the result. If the sign "-" are both members, therefore, you perform an addition and the result is also written with the sign "-". 3
In the presence of fractional values as coefficients of a polynomial, give to simplify the sample fractions to a common denominator. To do this, multiply all the coefficients of the expression for the same number so that when reducing fractions, leaving only the integer part. In the simplest case, the common denominator is the product of all denominators in fractional coefficients. After multiplying all the members, guide the facilitation of such events. 4
After bringing to a common denominator and combine like terms take out a common subexpression in parentheses. To do this, define the group members, where has the same subexpression. Divide the coefficients of the group on the General part and put it in front of the brackets. In parentheses Express not the whole polynomial, namely the group members with the remainder from dividing the coefficients. 5
Don't lose the sign during the removal of the brackets. If you want the common part to stand with the sign "-", then for each member in brackets will change the sign to opposite. The other members not parties to the removal of the brackets, write before or after the parentheses, preserving their sign.
6
If the brackets imposed a General part of the degree, for the group in parenthesis is a subtraction of index unbearable degree. When opening brackets extent such members are added and coefficients are multiplied.
7
The expression can be reduced to an integer, if it divided all the coefficients of the polynomial. Check, or in the given example, the common divisor. To do this, locate all of the coefficients of the number that evenly share each of them. Follow division all coefficients of the polynomial. 8
If the example is set to a literal of the variable, substitute it in the converted expression. Count and record the result. An example is solved.

# Advice 3 : How to learn to solve equations

An equation is a mathematical record of equality with one or more arguments. The solution of the equation consists in finding unknown values of arguments – roots, in which the specified equality is true. The equations are algebraic, algebraicheskie, linear, square, cubic etc. To solve them you need to learn the identity transformation, shifts, lookups and other operations to simplify the expression, maintaining a given equality. Instruction
1
Linear equation in General case has the form: ax + b = 0, and the unknown variable x here can be only in the first degree, she also should not be in the denominator. However, the problem is often the equation appears, for example, like this: x+2/4 + x = 3 – 2*x. In this case, before computing the argument it is necessary to bring the equation to the common view. This is accomplished by a series of transformations.
2
Drag a second (right) part of the equation on the other side of equality. Each term will change its sign: x+2/4 + x - 3 + 2*x = 0. Swipe the addition of arguments and numbers, simplify the expression: 4*x – 5/2 = 0. Thus, the obtained General form write linear equations, it is easy to find x: 4*x = 5/2, x = 5/8.
3
In addition to the operations described above, when solving equations, you should use 1 and 2 the identity transformation. Their essence lies in the fact that both parts of the equation can be folded with the same or multiply the same number or expression. The resulting equation will look different, but its roots will remain unchanged.
4
The solution of quadratic equations of the form ах2 + bх +C = 0 reduces to determining the coefficients a, b, C and their substitution in known formulas. As a rule, to give a General view you must do the transformation and simplification of expressions. Thus, the equation of the form x2 = (6x + 8)/2 open the brackets, leaving the right of the equal sign. Get the following entry: -x2 - 3x + 4 = 0. Multiply both parts of the equality by -1 and note the result: x2 + 3x - 4 = 0.
5
Calculate the discriminant of the square equation by the formula D = b2 – 4*a*c = 32 – 4*1*(-4) = 25. In case of positive discriminant the equation has two roots, formula of finding of which are: x1 = -b + √(D)/2*a; x2 = -b - √(D)/2*a. Substitute values and calculate: x1 = (-3+5)/2 = 1 x2 = (-3-5)/2 = -4. If the resulting discriminant is equal to zero, the equation would have only one root, it follows from the formulas for D
6
When finding the roots of cubic equations using the method of vieta-Cardano. More complex equations of degree 4 are calculated using the replacement, which decreases the degree of the arguments and equations are solved in several stages, as the square.
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