The solution is $$(4,2)$$:  $$j=4$$ and $$d=2$$. Answer Key provided. This resource works well as independent practice, homework, extra Multiplying and Dividing, including GCF and LCM, Powers, Exponents, Radicals (Roots), and Scientific Notation, Introduction to Statistics and Probability, Types of Numbers and Algebraic Properties, Coordinate System and Graphing Lines including Inequalities, Direct, Inverse, Joint and Combined Variation, Introduction to the Graphing Display Calculator (GDC), Systems of Linear Equations and Word Problems, Algebraic Functions, including Domain and Range, Scatter Plots, Correlation, and Regression, Solving Quadratics by Factoring and Completing the Square, Solving Absolute Value Equations and Inequalities, Solving Radical Equations and Inequalities, Advanced Functions: Compositions, Even and Odd, and Extrema, The Matrix and Solving Systems with Matrices, Rational Functions, Equations and Inequalities, Graphing Rational Functions, including Asymptotes, Graphing and Finding Roots of Polynomial Functions, Solving Systems using Reduced Row Echelon Form, Conics: Circles, Parabolas, Ellipses, and Hyperbolas, Linear and Angular Speeds, Area of Sectors, and Length of Arcs, Law of Sines and Cosines, and Areas of Triangles, Introduction to Calculus and Study Guides, Basic Differentiation Rules: Constant, Power, Product, Quotient and Trig Rules, Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change, Implicit Differentiation and Related Rates, Differentials, Linear Approximation and Error Propagation, Exponential and Logarithmic Differentiation, Derivatives and Integrals of Inverse Trig Functions, Antiderivatives and Indefinite Integration, including Trig Integration, Riemann Sums and Area by Limit Definition, Applications of Integration: Area and Volume, You’re going to the mall with your friends and you have, Wouldn’t it be clever to find out how many pairs of jeans and how many dresses you can buy so you use the whole, (Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the, Wouldn’t it be clever to find out how many pairs of jeans and how many dresses you can buy with your, Let’s say at the same store, they also had pairs of shoes for, Now we have a new problem: to spend the even, $$\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$$, $$\displaystyle \begin{array}{l}5x-6y-\,7z\,=\,\,7\\6x-4y+10z=\,-34\\2x+4y-\,3z\,=\,29\,\end{array}$$, $$\displaystyle \begin{array}{l}6x-4y+10z=-34\\\underline{{2x+4y-\,3z\,=\,29}}\\8x\,\,\,\,\,\,\,\,\,\,\,\,\,+7z=-5\end{array}$$, The totally yearly investment income (interest) is, How many liters of these two different kinds of milk are to be mixed together to produce, A store sells two different types of coffee beans; the more expensive one sells for, The beans are mixed to provide a mixture of, How much of each type of coffee bean should be used to create, minutes earlier than Megan (we have to put minutes into hours by dividing by. Simultaneous equations (Systems of linear equations): Problems with Solutions. Solve for $$d$$: $$\displaystyle d=-j+6$$. $$\displaystyle \begin{array}{l}\color{#800000}{{2x+5y=-1}}\,\,\,\,\,\,\,\text{multiply by}-3\\\color{#800000}{{7x+3y=11}}\text{ }\,\,\,\,\,\,\,\text{multiply by }5\end{array}$$, $$\displaystyle \begin{array}{l}-6x-15y=3\,\\\,\underline{{35x+15y=55}}\text{ }\\\,29x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=58\\\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\2(2)+5y=-1\\\,\,\,\,\,\,4+5y=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5y=-5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-1\end{array}$$. Graph each equation on the same graph. $$\begin{array}{c}L=M+\frac{1}{6};\,\,\,\,\,\,5L=15M\\5\left( {M+\frac{1}{6}} \right)=15M\\5M+\frac{5}{6}=15M\\30M+5=90M\\60M=5;\,\,\,\,\,\,M=\frac{5}{{60}}\,\,\text{hr}\text{. In algebra, a system of equations is a group of two or more equations that contain the same set of variables. This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself. Understand these problems, and practice, practice, practice! If the equation is written in standard form, you can either find the x and y intercepts or rewrite the equation in slope intercept form. (Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the Exponents and Radicals in Algebra section.). How much did she invest in each rate? Substitution is the favorite way to solve for many students! The trick to do these problems “by hand” is to keep working on the equations using either substitution or elimination until we get the answers. Word problem using system of equations (investment-interest) Example: A woman invests a total of 20,000 in two accounts, one paying 5% and another paying 8% simple interest per year. Graphing Systems of Equations Practice Problems. Solution : Let "x" be the number. The cool thing is to solve for 2 variables, you typically need 2 equations, to solve for 3 variables, you need 3 equations, and so on. When equations have no solutions, they are called inconsistent equations, since we can never get a solution. Here is a set of practice problems to accompany the Linear Systems with Two Variables section of the Systems of Equations chapter of the notes … Add 18 to both sides. If bob bought six items for a total of 18, how many did he buy of each? We’ll need to put these equations into the \(y=mx+b$$ ($$d=mj+b$$) format, by solving for the $$d$$ (which is like the $$y$$): $$\displaystyle j+d=6;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+6\text{ }$$, $$\displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{50}}=-\frac{1}{2}j+4$$. This will give us the two equations. Let’s go for it and solve:     $$\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$$: $$\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+50d+20s=260\\j=2s\end{array}$$, \displaystyle \begin{align}2s+d+s&=10\\25(2s)+50d+\,20s&=260\\70s+50d&=260\end{align}, $$\displaystyle \begin{array}{l}-150s-50d=-500\\\,\,\,\,\,\underline{{\,\,70s+50d=\,\,\,\,260}}\\\,\,-80s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-240\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s=3\\\\3(3)+d=10;\,\,\,\,\,d=1\,\\j=2s=2(3);\,\,\,\,\,\,j=6\end{array}$$. Use easier numbers if you need to: if you buy. Maybe the problem will just “work out” so we can solve it; let’s try and see. $$\require {cancel} \displaystyle \begin{array}{c}10\left( {8w+12g} \right)=1\text{ or }8w+12g=\frac{1}{{10}}\\\,14\left( {6w+8g} \right)=1\text{ or }\,6w+8g=\frac{1}{{14}}\end{array}$$, $$\displaystyle \begin{array}{c}\text{Use elimination:}\\\left( {-6} \right)\left( {8w+12g} \right)=\frac{1}{{10}}\left( {-6} \right)\\\left( 8 \right)\left( {6w+8g} \right)=\frac{1}{{14}}\left( 8 \right)\\\cancel{{-48w}}-72g=-\frac{3}{5}\\\cancel{{48w}}+64g=\frac{4}{7}\,\\\,-8g=-\frac{1}{{35}};\,\,\,\,\,g=\frac{1}{{280}}\end{array}$$             $$\begin{array}{c}\text{Substitute in first equation to get }w:\\\,10\left( {8w+12\cdot \frac{1}{{280}}} \right)=1\\\,80w+\frac{{120}}{{280}}=1;\,\,\,\,\,\,w=\frac{1}{{140}}\\g=\frac{1}{{280}};\,\,\,\,\,\,\,\,\,\,\,w=\frac{1}{{140}}\end{array}$$. Systems of equations » Tips for entering queries. Difficult. The money spent depends on the plumber’s set up charge and number of hours, so let $$y=$$ the total cost of the plumber, and $$x=$$ the number of hours of labor. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. Use substitution and put $$r$$ from the middle equation in the other equations. On to Algebraic Functions, including Domain and Range – you’re ready! $$\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}37x+4y=124\,\\x=4\,\end{array}}\\\\37(4)+4y=124\\4y=124-148\\4y=-24\\y=-6\end{array}$$. To start, we need to define what we mean by a linear equation. Problem 1. Always write down what your variables will be: Let $$j=$$ the number of jeans you will buy, Let $$d=$$ the number of dresses you’ll buy. It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation. How far is the mall from the sisters’ house? $$\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}j+d=\text{ }6\\25j+50d=200\end{array}}\\\\25j+50(-j+6)=200\\25j-50j+300=200\\-25j=-100\,\,\\j=4\,\\d=-j+6=-4+6=2\end{array}$$. See how similar this problem is to the one where we use percentages? Note that, in the graph, before 5 hours, the first plumber will be more expensive (because of the higher setup charge), but after the first 5 hours, the second plumber will be more expensive. by Visticious Loverial (Austria) The sum of four numbers a, b, c, and d is 68. Notice that the $$j$$ variable is just like the $$x$$ variable and the $$d$$ variable is just like the $$y$$. Let’s let $$j=$$ the number of pair of jeans, $$d=$$ the number of dresses, and $$s=$$ the number of pairs of shoes we should buy. Introduction and Summary; Solving by Addition and Subtraction; Problems; Solving using Matrices and Row Reduction; Problems ; Solving using Matrices and Cramer's Rule; Problems; Terms; Writing Help. Subjects: Algebra, Word Problems, Algebra 2. If you can answer two or three integer questions with the same effort as you can onequesti… OK, enough Geometry for now! Since they have at least one solution, they are also consistent. (This is the amount of money that the bank gives us for keeping our money there.) First of all, to graph, we had to either solve for the “$$y$$” value (“$$d$$” in our case) like we did above, or use the cover-up, or intercept method. Graph each equation on the same graph. Also – note that equations with three variables are represented by planes, not lines (you’ll learn about this in Geometry). You will never see more than one systems of equations question per test, if indeed you see one at all. Probably the most useful way to solve systems is using linear combination, or linear elimination. You will probably encounter some questions on the SAT Math exam that deal with systems of equations. Note that there’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section. The first trick in problems like this is to figure out what we want to know. Then, let’s substitute what we got for “$$d$$” into the next equation. Thus, the plumber would be chosen based on how many hours Michaela’s mom thinks the plumber will be there. $$\displaystyle \begin{array}{c}\,\,\,3\,\,=\,\,3\\\underline{{+4\,\,=\,\,4}}\\\,\,\,7\,\,=\,\,7\end{array}$$, $$\displaystyle \begin{array}{l}\,\,\,12\,=\,12\\\,\underline{{-8\,\,=\,\,\,8}}\\\,\,\,\,\,4\,\,=\,\,4\end{array}$$, $$\displaystyle \begin{array}{c}3\,\,=\,\,3\\4\times 3\,\,=\,\,4\times 3\\12\,\,=\,\,12\end{array}$$, $$\displaystyle \begin{array}{c}12\,\,=\,\,12\\\frac{{12}}{3}\,\,=\,\,\frac{{12}}{3}\\4\,\,=\,\,4\end{array}$$, $$\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}j+d=6\text{ }\\25j+50d=200\end{array}}\\\\\,\left( {-25} \right)\left( {j+d} \right)=\left( {-25} \right)6\text{ }\\\,\,\,\,-25j-25d\,=-150\,\\\,\,\,\,\,\underline{{25j+50d\,=\,200}}\text{ }\\\,\,\,0j+25d=\,50\\\\25d\,=\,50\\d=2\\\\d+j\,\,=\,\,6\\\,2+j=6\\j=4\end{array}$$, Since we need to eliminate a variable, we can multiply the first equation by, $$\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+50d+\,20s=260\\j=2s\end{array}$$. Sometimes, however, there are no solutions (when lines are parallel) or an infinite number of solutions (when the two lines are actually the same line, and one is just a “multiple” of the other) to a set of equations. In this type of problem, you would also have/need something like this: we want twice as many pairs of jeans as pairs of shoes. $$\displaystyle x+y=6\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,y=-x+6$$, $$\displaystyle 2x+2y=12\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,y=\frac{{-2x+12}}{2}=-x+6$$. Now we have a new problem: to spend the even $260, how many pairs of jeans, dresses, and pairs of shoes should we get if want say exactly 10 total items? When we substitute back in the sum $$\text{ }j+o+c+l$$, all in terms of $$j$$, our $$j$$’s actually cancel out, which is very unusual! You discover a store that has all jeans for$25 and all dresses for $50. “Systems of equations” just means that we are dealing with more than one equation and variable. $$x$$ plus $$y$$ must equal 180 degrees by definition, and also $$x=2y-30$$ (Remember the English-to-Math chart?) And we’ll learn much easier ways to do these types of problems. Let’s check our work: The two angles do in fact add up to 180°, and the larger angle (110°) is 30° less than twice the smaller (70°). Systems of Equations: Students will practice solving 14 systems of equations problems using the substitution method. Systems of Equations Word Problems Example: The sum of two numbers is 16. You’ll want to pick the variable that’s most easily solved for. Since we have the $$x$$ and the $$z$$, we can use any of the original equations to get the $$y$$. 30 Systems Of Linear Equations Word Problems Worksheet Project List. What is the value of x? If you missed this problem, review . At how many hours will the two companies charge the same amount of money? What we want to know is how many pairs of jeans we want to buy (let’s say “$$j$$”) and how many dresses we want to buy (let’s say “$$d$$”). We’ll substitute $$2s$$ for $$j$$ in the other two equations and then we’ll have 2 equations and 2 unknowns. This will actually make the problems easier! Find the solution n to the equation n + 2 = 6, Problem 2. Problem 2. Remember these are because of the Additive Property of Equality, Subtraction Property of Equality, Multiplicative Property of Equality, and Division Property of Equality:eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-2','ezslot_10',128,'0','0']));eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-2','ezslot_11',128,'0','1']));eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-2','ezslot_12',128,'0','2'])); So now if we have a set of 2 equations with 2 unknowns, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Now we use the 2 equations we’ve just created without the $$y$$’s and solve them just like a normal set of systems. Grades: 6 th, 7 th, 8 th, 9 th, 10 th, 11 th. If we increased b by 8, we get x. … The larger angle is 110°, and the smaller is 70°. For each correct answer to a math problem, you will enter a 30-second bonus round. Given : 18 is taken away from 8 times of the number is 30 Then, we have. Let’s do one more with three equations and three unknowns: She has$610 to spend (including tax) and wants 24 flowers for each bouquet. Define a variable, and look at what the problem is asking. When I look at this version, these two, this system of equations right over here on the left, where I've already solved for L, to me this feels like substitution might be really valuable. Il en résulte un système d'équations linéaire résolu en fonction des concentrations inconnues. Here’s one that’s a little tricky though: Let’s do a “work problem” that is typically seen when studying Rational Equations – fraction with variables in them –  and can be found here in the Rational Functions, Equations and Inequalities section. Find the slope and y-intercept of the line \ (3x ... we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines. Thereby, a resultant linear equation system is solved as a function of the unknown concentrations. Set the distances together, since the two sisters live the same distance from the mall. We also could have set up this problem with a table: How many liters of these two different kinds of milk are to be mixed together to produce 10 liters of low-fat milk, which has 2% butterfat? Note that there’s an example of a Parametric Distance Problem here in the Parametric Equations section. This one is actually easier: we already know that $$x=4$$. 8x - 18 = 30. Wow! Solving for $$x$$, we get $$x=2$$. Problem 3. One number is 4 less than 3 times … Types: Activities, Games, Task Cards. The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. If we decrease c by 15, we get 2x.If we multiply d by 4, we get x. Now we can plug in that value in either original equation (use the easiest!) eval(ez_write_tag([[300,250],'shelovesmath_com-leader-2','ezslot_14',134,'0','0']));eval(ez_write_tag([[300,250],'shelovesmath_com-leader-2','ezslot_15',134,'0','1']));eval(ez_write_tag([[300,250],'shelovesmath_com-leader-2','ezslot_16',134,'0','2']));Here’s another problem where we’re trying to compare two different scenarios. Show Step-by-step Solutions. 8x - 18 = 30 After “pushing through” (distributing) the 5, we multiply both sides by 6 to get rid of the fractions. Here’s one like that: She then buys 1 pound of jelly beans and 4 pounds of caramels for $3.00. Consistent: If a system of linear equations has at least one solution, then it is called consistent. First, we get that $$s=3$$, so then we can substitute this in one of the 2 equations we’re working with. Below are our two equations, and let’s solve for “$$d$$” in terms of “$$j$$” in the first equation. You really, really want to take home 6 items of clothing because you “need” that many new things. Solving Systems with Linear Combination or Elimination, If you add up the pairs of jeans and dresses, you want to come up with, This one’s a little trickier. Algebra Solving Age Problems Using System of Equations - Duration: 23:11. We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes. Many systems of equations word problem questions are easy to confuse with other types of problems, like single variable equations or equations that require you to find alternate expressions. The point of intersection is the solution to the system of equations. This math worksheet was created on 2013-02-14 and has been viewed 18 times this week and 2,037 times this month. Solution … solving system of linear equations by substitution y=2x x+y=21 Replace y = 2x into the second equation. Pretty cool! There are some examples of systems of inequality here in the Linear Inequalities section. Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In the following practice questions, you’re given the system of equations, and you have to find the value of the variables x and y. Ron Woldoff is the founder of National Test Prep, where he helps students prepare for the SAT, GMAT, and GRE. Let’s say at the same store, they also had pairs of shoes for$20 and we managed to get $60 more from our parents since our parents are so great! Displaying top 8 worksheets found for - Systems Of Equations Problems. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life!). Marta Rosener 3,154 views. But we can see that the total cost to buy 1 pound of each of the candies is$2. We could buy 4 pairs of jeans and 2 dresses. From counting through calculus, making math make sense! Age word problems. See – these are getting easier! She wants to have twice as many roses as the other 2 flowers combined in each bouquet. Easy. How much of each type of coffee bean should be used to create 50 pounds of the mixture? Solve, using substitution: $$\displaystyle \begin{array}{c}x+y=180\\x=2y-30\end{array}$$, $$\displaystyle \begin{array}{c}2y-30+y=180\\3y=210;\,\,\,\,\,\,\,\,y=70\\x=2\left( {70} \right)-30=110\end{array}$$. You may need to hit “ZOOM 6” (ZoomStandard) and/or “ZOOM 0” (ZoomFit) to make sure you see the lines crossing in the graph. You really, really want to take home 6 items of clothing because you “need” that many new things. In these cases, the initial charge will be the $$\boldsymbol {y}$$-intercept, and the rate will be the slope. Math exercises for everyone. Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination: \displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}, $$\displaystyle \begin{array}{l}5x-6y-\,7z\,=\,\,7\\6x-4y+10z=\,-34\\2x+4y-\,3z\,=\,29\,\end{array}$$    $$\displaystyle \begin{array}{l}6x-4y+10z=-34\\\underline{{2x+4y-\,3z\,=\,29}}\\8x\,\,\,\,\,\,\,\,\,\,\,\,\,+7z=-5\end{array}$$, $$\require{cancel} \displaystyle \begin{array}{l}\cancel{{5x-6y-7z=7}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,20x-24y-28z\,=\,28\,\\\cancel{{2x+4y-\,3z\,=29\,\,}}\,\,\,\,\,\,\,\,\underline{{12x+24y-18z=174}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,32x\,\,\,\,\,\,\,\,\,\,\,\,\,\,-46z=202\end{array}$$, $$\displaystyle \begin{array}{l}\,\,\,\cancel{{8x\,\,\,+7z=\,-5}}\,\,\,\,\,-32x\,-28z=\,20\\32x\,-46z=202\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,32x\,-46z=202}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-74z=222\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z=-3\end{array}$$, $$\displaystyle \begin{array}{l}32x-46(-3)=202\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{202-138}}{{32}}=\frac{{64}}{{32}}=2\\\\5(2)-6y-\,\,7(-3)\,=\,\,7\,\,\,\,\,\,\,\,y=\frac{{-10+-21+7}}{{-6}}=4\end{array}$$. We add up the terms inside the box, and then multiply the amounts in the boxes by the percentages above the boxes, and then add across. We can also write the solution as $$(x,-x+6)$$. Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: … I know – this is really difficult stuff! Notice that the slope of these two equations is the same, but the $$y$$-intercepts are different. Graphs of systems of equations are really important because they help model real world problems. So far we’ll have the following equations: $$\displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}$$. Here’s a distance word problem using systems. How many roses, tulips, and lilies are in each bouquet? 2 fancy shirts and 5 plain shirts 2) There are 13 animals in the barn. Then, we have. Which is the number? WORD PROBLEMS ON SIMPLE EQUATIONS. So far, we’ve basically just played around with the equation for a line, which is $$y=mx+b$$. Now you should see “Second curve?” and then press ENTER again. 23:11 . And if we up with something like this, it means there are no solutions: $$5=2$$  (variables are gone and two numbers are left and they don’t equal each other).

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