i have no idea at all how to do this. for example one of the questions is y=6x-11, -2x-3y= -7 i have no clue how to solve this. i think this is useless but can someone please help me?!?!
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Algebrator can easily solve problems such as the one you posted. It´s a great tool in order to rapidly complete Algebra assignments.
Your problems have been solved using Algebrator as shown in the link below. Just copy-paste the url below on your browser. You will see the problems solved “step-by-step” and also an explanation of each step. We hope you find it useful!
1. In order to create an equation containing only one variable, we need to subtitute the right side of the equation for Y in the second equation. In our example, all instances of Y in the second equation will be replaced by 6x-11.
y = 6x-11
-2x - 3(6x-11) = -7
2. We need to expand this term by multiplying a term and an expression.
The following product distributive property will be used: A(B+C) = AB+AC
In our example, the resulting expression will consist of 2 terms: the first term is a product of 3 and 6x and the second term is a product of 3 and -11.
y = 6x-11
-2x - (3.6x - 3.11) = -7
3. Numerical factors are then multiplied:
y = 6x-11
-2x - (18x - 33) = -7
4. We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged.
In our example, the following 2 terms will change sign: 18x and -33.
y = 6x-11
-2x + 18x + 33 = -7
5. We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1.
There is only one group of like terms: -2x, -18x
y = 6x-11
-20x + 33 = -7
6. In order to solve this linear equation, we need to group all the variable terms on one side, and all the constant terms on the other side of the equation.
In our example, - term 33, will be moved to the right side.Notice that a term changes sign when it ‘moves’ from one side of the equation to the other.
y = 6x-11
-20x = -40
7. In order to isolate the variable in this linear equation, we need to get rid of the coefficient that multiplies it. This can be accomplished if both sides are divided by -20.
y = 6x-11
x = 2
8. In order to eliminate x in the first equation, we need to substitute the right side of the second equation for x in the first equation. In our example, all instances of x in the first equation will be replaced by: 2
y = 6.2 - 11
x = 2
9. The given system of equations is consistent and independent.
x = 2
y = 1
10. Following the link below you will also see the graph for this System.
It is important to understand that anytime you are solving a system of equations you are answering the basic question “Do the graphs of these equation touch?”. They may touch in one point (one solution), no points (no solutions) or an infinite number of points (i.e. the equations define the same line).
.
y = 6x - 11
-2x - 3y = -7
One of the equations is solve for y, which is equal to 6x - 11. This means that the y in the second equation is also equal to 6x - 11 since they are the same variable, so substitute it and solve for x.
-2x - 3y = -7 (substitute 6x - 11 for y)
-2x - 3(6x - 11) = -7 (distribute)
-2x - 18x + 33 = -7 (combine like terms)
-20x + 33 = -7 (subtract 33 from both sides)
-20x = -40 (divide both sides by -20)
x = 2
Now go back to either of the equations to solve for y. Since one equation is already solved for y, it will be easiest to use that equation.
y = 6x - 11 (substitute 2 for x)
y = 6(2) - 11
y = 12 - 11
y = 1
ANSWER: x = 2, y = 1; point (2, 1)
yeah since y = 6x - 11 plug 6x - 11 into -2x-3y= -7 so you get -2x -3(6x - 11) =-7
solve for x
x = 2
then plug 2 into y = 6x - 11
y = 6(2) - 11
y = 1