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Solved the system by substitution.if the system is inconsistent and has no solution,state this .if the system is dependent,write the form of the solution for any real number x.
(1) x=4y+3
(2) x=-2/3y+3/4

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Isaak B. | Good (H.S. or College Math, Physics, Chem, EE Engineering) Cheap TutorGood (H.S. or College Math, Physics, Che...
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The idea of solving a system by substitution is to eliminate the unknowns by converting some of the equations into a form which can be substituted into the other equation.   To do this, you have to:
1) Decide which variable to eliminate and which to solve for (At the end of the problem you can go back and solve for the other variable if the question asks you to).  It doesn't matter which if you perform correct operations on each equation.
2) Solve at least one of the equations for the variable you want to solve for.  In other words, isolate that variable so it appears all alone on one side of the equation only, the way y is in your third problem (in both equations).
3) substitute the expression that your solved equation shows that your isolated variable is equal to into another equation every place it appears in the other equation, then simplify.  This gives you an equation in which the variable you solved for has been eliminated.
4) Repeat steps 2) and 3) until you end up with an equation with only one variable.
5) Solve the equation with only one variable in it for that variable.  
6 (if necessary per the problem requirements) substitute your result for that variable back into the other equations to solve for the other variables.
I'll work the first one, then you try the rest.
1) x = 4y  + 3  [A]
    2x + 5y = -7 [B]
The capital letters are my labels for the equations.
From equation [A], we see that x is already by itself.  So let's use that as the result of what I called step 2, and decide (step 1) to eliminate x first in this problem.  That way we are already on step 3, because step 2 won't be necessary in this case.
Substituting the right hand side of [A] into [B] wherever "x" appears, because equation [A] tells us that for all points (x,y) at which equation [A] is true, x is equal to the right hand side of [A].
This gives us : 
2(4y+3) + 5y = -7  (step 4 is now complete!)
Simplifying, you can now complete step 5.
Ok, at that point you will know the y-coordinate of the location(s) at which equation [A] and equation [B] are both true.  Plug this result back into [A] or [B] to get the final answer.  To check your work, plug your answers into both equations and verify that they are both true statements using your solution in place of the variables.
2) For this question, you will find that solving the equations gives you a result that cannot be true.  What does this tell you?  What do you notice about the two equations if you manipulate them so they are each in the form y = mx + b? (solve each for y)  How would a graph of these lines look?  Where is the intercept (if any?)
3)  These lines are already in the form y = mx + b.  What does that tell you about their graphs? How would that look?  Sketch it out if you can't see it in your mind.
If each equation describes all the (x,y) points at which their respective lines have points, can you combine the equations to identify the x coordinate ?   If y = 5x+2 for all of the first line points, and y = 9x-6 for all of the second line points, for what point or points would 5x+2 = 9x-6?  Focus on the similarities, don't let the specific expression throw you.  Said another way, abstracting the information if y1="blah" for all of the first line points, and y = "fooey" for all of the second line points, for what point or points would blah = fooey?  The intersection!   But if 5x+2 = 9x-6 then what is x?  
4) I think if you work through the above, you can do this.  If not, tell me more about what the problem is. Show me your work.
What they didn't show you is how to handle a dependent system.  
Here's an example:
6x + 3y = 12 [G]
12x + 6y = 24 [H]
Here, as in all dependent systems of equations, one of the equations is just a different version of the other.  In our specific case, [H] = 2* [G].   It is not always as obvious as this, that one equation is dependent on the other, so we'll pretend we don't notice this yet so we can see what typically happens when you go ahead and apply the usual approaches (substitution or graphical solution-finding, for instance) to such a system.
Suppose we go ahead and put both equations in the form y = mx + b.  If we simplify both results, we'd get y = -2x+4, but even if we didn't simplify both to this form, our graphs of the lines of these would lie on top of one another.  In other words, everywhere along either line is also on the other line, or the whole line is a solution.
If we had decided to use substitution, we may have solved the first expression for y and got y=-2x+4, but when we substitute that into the other equation:
12x+6(-2x+4) = 24
0x + 24 = 24
Uh-oh, no matter what x is, the equation is true.  We can simplify this to:
0 = 0.
This is another form of the equation (still true no matter what x is).  This means we just let x be x, and write one of the other equations that has the other variable, and x, in it, sovled for the other variable so we can describe how the solution depends on x, even if there is no particular, unique, solution to be found.
y = -2x +4 is such a description.