Systems of Linear Equations. How to: solve by graphing, substitution, linear combinations, and special types of linear systems By: Sarah R. Algebra 1; E block. What is a Linear System, Anyways?. A linear system includes two, or more, equations, and each includes two or more variables.
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How to: solve by graphing, substitution, linear combinations, and special types of linear systems
By: Sarah R.
Algebra 1; E block
1. Solve one equation for one of its variables
2. Substitute that expression into the other equation and solve for the other variable
3. Substitute that value into first equation; solve
4. Check the solution
See next page for a step by step example!Solving Linear Systems by Substitution
Here’s the problem: Equation one -x+y=1 Equation two 2x+y=-2
Try this on your own…but if you need help or a few pointers…see the next page!Example: The Substitution Method
First, solve equation one for y two 2x+y=-2
Next, substitute the above expression in for “y” in equation two, and solve for x
Substitute “x+1” for y
simplify the above expression
Subtract one from both sides (because your goal is to solve for x)
Solve for x ( divide both sides by 3; since x is being multiplied by three, and you need it alone, so do the inverse operation: divide by 3)
Congratulations! You now know x has a value of –1…but you still need to find “y”.
To do so…
First, write down equation one
Substitute –1 for x, since you just found that x=-1
Solve the equation for y by adding –1 +1
So, now what?
You’re done; simply write out the solution as (-1,0)
***Did you remember??
To write a solution, once you’ve found x and y, you must put x first and then y: (x,y)
Equation Two: 3x-2y=11
You should solve for y in the first equation. Again, you lessen your work because there is no leading coefficient before the y in equation one, while there are leading coefficients with all the other variables.
2. After looking at the coefficients of x and y, you need to multiply one or both equations by a number that will give you new coefficients for x or y that are opposites.
3. Add the equations and solve for the unknown variable
4. Substitute the value gotten in step 3 into either of the original equations; solve for other variable
5. Check the solution in both original equations
Here’s the original problem: two 2x+y=-2
Solve the linear system
Equation 1: 3x+5y=6
Equation 2: -4x+2y=5
Do you remember the first step?
…put the equations into columns
Now, you need to multiply each equation by a number that will cause your leading coefficients of either x or y to become opposites. In this case, try to get opposite coefficients for x. to do this, multiply the first equation by four and the second by three.
***You must multiply all terms by 3 or 4
3x+5y=6, when all terms are multiplied by four, this equation will be: 12x+20y=24
-4x+2y=5, when all terms are multiplied by three, this equation will be: -12x +6y=15
Your next step is to add the two revised equations:
+ (-12x) + 6y= 15
26y=39 (sum of equations)
To get the “y” alone, you must divide each side by 26, (you divide since the y is being multiplied by 26, and to isolate the y you do the inverse operation)
So, you have found “Y”, but you aren't done yet!
What’s left, you may be thinking…well, you have only found “y”…what about x?
To find x, you have to place “y” into equation 2.
Equation 2: -4x+2y=5
Substitute the value you just found for “y” : 3
simplify by multiplying 2 by three-halves
subtract 3 from both sides because you are working to isolate x
solve for x by dividing both sides by –4 (inverse operation)
The solution to the example system is (-1, 3)
Graph and Check
Here’s the problem: Equation one x+y=(-2) Equation two 2x-3y=(-9)
Try this problem out…but a step by step process follows!An Example of the Graph and Check Method
Equation one: originally, it was: x+y=(-2) but after putting it into slope intercept, it reads: y=(-x)-2
Equation two: originally, it was: 2x-3y=(-9), but once in slope intercept, it reads: y= 2x+3
From the above equations, you can make the following conclusions:
Equation one has a slope of –1 and a y intercept of –2
Equation two has a slope of 2 and a y intercept of 3
***Remember that in the slope intercept form (y=mx+b), m is the slope; b is the y intercept
now, you will be able to graph the two equations as lines. Once done this, you can conclude that the lines seem to intercept at (-3,1).
To check this assumption, put (-3) in for x and 1 in for y in BOTH EQUATIONS, and solve both:
Equation one: (-3)+(-1)=-2
Equation two: 2(-3)-3(1)=-6-3=-9
Since both equations, once solved, equaled what they should have, you know that the solution to this linear system is (-3,1)
Special types of Linear Systems
1. The boundary line on the graph will be dashed if the inequality is < or >.
2. The boundary line will be solid if the inequality is <or >.
3. You will also notice that graphs of linear inequalities are shaded in certain areas. To decide where to shade, pick a point that is CLEARLY above the line, and a point that is CLEARLY below the line. Put the first point into the inequality; solve; then do the same for the other point. Whichever point works, you shade that side.
*both will be horizontal lines because there is no x whatsoever in either equation
Now, you can graph the equation (next page)
Equation one: -2x+3y=6
Equation two: 2x+y=10
Equation one: x-6y=-19
Equation two: 3x-2y=-9
Equation one: x+3y=7
Equation two: 4x-7y=-10
See next page for more; answers on last page
Equation one: -2x-3y=4
Equation two: 2x-4y=3
Equation one: 3x-5y=-4
Equation two: -9x+7y=8
The End form