**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. • When two equations are used to model a problem, it is called a linearsystem.

**Before You Begin…Important Terms to know** • Linear system: two equations that form one equation • Solution: the answer to a system of linear equation; must satisfy both equations ***: a solution is written as an ordered pair: (x,y) • Leading Coefficient: any given number that is before any given variable (for example, the leading coefficient in 3x is 3.) • Isolate: to get alone

**Ways to Solve Linear Systems** By: Substitution

**Basic steps:** 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** Y=x+1 Next, substitute the above expression in for “y” in equation two, and solve for x Here’s how: Equation two 2x+y=-2 Substitute “x+1” for y 2x+ (x+1)=-2 simplify the above expression 3x+1=-2 Subtract one from both sides (because your goal is to solve for x) 3x=-3 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) X=-1 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 Y=x+1 Substitute –1 for x, since you just found that x=-1 Y= (-1)+1 Solve the equation for y by adding –1 +1 Y=0 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)

**Things to Know About the Substitution Method** • 1. It doesn’t matter if you choose to solve for y or x first; the answer or solution will be the same either way. • 2. You can also choose to solve equation two before equation one; simply follow the same steps, just using a slightly rearranged order. • *** You should always decide whether to solve x or y first, or equation one or two first, depending on which way is more efficient (See next page!)

**Deciding the Order in Which to Solve** • Here is an instance where it is easier to solve equation two first (for x) Equation One: 3x-2y=1 Equation Two: x+4y=3 By solving equation two first, you are lessening your work, because there is no leading coefficient before the x in equation two, so you don’t have to worry about dividing to isolate the “x” • Here is an instance where you help yourself by solving equation one for y Equation One: 2x+y=5 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.

**Solving Linear Systems by Linear Combinations**

**Solving Systems by means of Linear Combinations** • Basic steps: 1. Arrange the equations with like terms in columns 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

**Example: Solving Systems by Linear Combinations** • Here’s an example…try it out, but if you have any problems, see the next page for a guided, step by step explanation • Solve this linear equation: Equation One: 3x+5y=6 Equation Two: -4x+2y=5

**Here’s the original problem:** 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 3x+5y=6 -4x+2y=5 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+ 20y=24 + (-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 2 -4x+2(3)=5 2 simplify by multiplying 2 by three-halves -4x+3=5 subtract 3 from both sides because you are working to isolate x -4x=2 solve for x by dividing both sides by –4 (inverse operation) x=-1 2 The solution to the example system is (-1, 3) 2 2

**A Final way to Solve Systems:** Graph and Check

**Here’s a method called graph and check** • Basic steps: 1. Put each equation into slope intercept form (y=Mx+B) 2. Graph the two lines (M is your slope; B is your Y-intercept) 3. Find the point that the lines appear to intersect at, and then put that solution into EACH equation and solve to check for accuracy.

**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

**The first step is to put the equations into slope-intercept** form 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 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 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)

**Don’t Let These Fool You…** Special types of Linear Systems

**Linear Systems with NO Solution** • Here’s the problem: Equation one: 2x+y=3 Equation two: 4x+2y=8 • After trying the graph method, you’ll find that the lines are parallel( don’t intersect) and therefore have no solution • After trying either of the substitution or linear combination methods, you will have an equation that cannot be dealt with. You will know that this is the case because it will make no senses whatsoever. Therefore, you have no solution to the system.

**Linear System with MANY Solutions** • If you use the graph method, you will see that the equations are the same line, and any point on the line is a solution. • If you use linear combinations or substitution, you will have a number =number, but both numbers will be the same. For example, 7=7 or 1=1. This indicates that the systems has many solutions.

**Solving Systems of Linear Inequalities**

**Graphing Systems of Linear Inequalities** • Here are some pointers and things to know: 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.

**An Example of Graphing Linear Inequalities** • Y<4 • Y>1 • Try this one out! Remember the steps; you can always go back a page if necessary…or go forward one page to get step by step guidance.

**So, You Needed Help…** • Here’s the original problem: y<4, y>1 • First, make a few basic conclusions: * The line for both boundaries will be dotted or dashed because it is < or >. *both will be horizontal lines because there is no x whatsoever in either equation Now, you can graph the equation (next page)

**Graphing Errors** • If your graph looked like the previous slide, you can congratulate yourself on getting the lines drawn correctly. However you forgot to: • Label the axis • Label the lines • Pick points and follow the previously described process to find where to shade (between y=1 and y=4) • Write the equation on the line

**In Simpler Terms: Graphing Systems of Linear Inequalities** • 1. Sketch the lines of each inequality (remember to use dashed lines for < or > and solid lines for < or >) • 2. LIGHTLY SHADE the area that is found by choosing points and placing them into the equation • 3. The final result, or answer, is the area that is where the shaded planes intersect, for example, in the previous problem, the answer is anywhere between the boundary lines of y=4 and y=1.

**To Make it Somewhat Easier…** • Basic guidelines for linear systems: • 1. Use the graphing method to get an approximate answer, to check a solution, or to give a visual idea of the system • 2. Using substitution or linear combinations will allow you to get an exact and more accurate answer • 3. Substitution helps a lot when used in systems that have coefficients of 1 or –1. • 4. When there isn't a 1 or –1 as coefficients, the linear combinations method is efficient.

**Fun, Fun: Examples to do on Your Own(Answers are on Last** Page) • 1. Solve the following Linear System by graphing Equation one: -2x+3y=6 Equation two: 2x+y=10 • 2. Solve the following Linear system by means of substitution Equation one: x-6y=-19 Equation two: 3x-2y=-9 • 3. Solve the following Linear system by means of substitution Equation one: x+3y=7 Equation two: 4x-7y=-10 See next page for more; answers on last page

** A Little More Fun: More Examples** • 4. Use linear combinations to solve this system Equation one: -2x-3y=4 Equation two: 2x-4y=3 • 5. Use linear combinations to solve this system Equation one: 3x-5y=-4 Equation two: -9x+7y=8

**Answers to the Examples** • 1. Your graph should show a point of intersection, which is your solution, of (3,4). • 2. (-1,3) • 3. (1,2) • 4. (-1.5,9) • 5. (-.5, .5)

**The End**