Additional Examples of using the Elimination Method to Solve Systems of Equations

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1 Additional Examples of using the Elimination Method to Solve Systems of Equations. Adjusting Coecients and Avoiding Fractions To use one equation to eliminate a variable, you multiply both sides of that equation by: Notice that this is a fraction. Example: to solve the system coecient of variable in other equation coecient of variable in equation being multiplied 5x + y 8 7x 8y 38 we could eliminate x by multiplying the rst equation by 7 5, or by multiplying the second equation by 5 7. We could eliminate y by multiplying the rst equation by or by multiplying the second equation by 8 4 3, If you are uncomfortable with fractions (though do be aware that they can show up as solutions even if the coecients are whole numbers), try this alternate start for the elimination method:. Pick a variable to eliminate. Make sure that variable appears once in each equation. 2. Multiply both sides of the top equation by the coecient of that variable in the bottom equation. 3. Multiply both sides the bottom equation by the original coecient of that variable in the top equation. 4. Negate both sides of either the top equation, or the bottom one, but not both.

2 Let us eliminate x from 5x + y 8 7x 8y 38 Multiply the top equation by 7, the coecient of x in the bottom equation: 35x + 42y 5 7x 8y 38 Multiply the bottom equation by 5, the original coecient of x in the top equation: 35x + 42y 5 35x 40y 90 Negate one, but not both, of the equations. Let us negate the top equation so that its right hand side is positive. This will make it easier to add the equations: Simplifying: 35x 42y ( 5) 35x 40y 90 35x 42y 5 35x 40y 90 Notice that the coecients of x are additive inverses. We may proceed with the rest of the elimination method. 82y 24. Solve for y: y Plug y 3 into an original equation to nd x: 5x + ( 3) 8 5x 8 8 5x 0 x 2 The solution is x 2, y 3. Let us check our work. Plug x 2, y 3, into both equations, and see if we get the same number on each side: 5(2) + ( 3)

3 7(2) 8( 3) Hence x 2, y 4 is a solution to the system. You may want to practice this by solving this system, and eliminate y. See if you get the same answers as we did when we eliminated x. 3

4 2. Eliminating a Power of a Variable Let us solve x 2 + y 2 7 2x 2y 2 We cannot eliminate x or x 2 with each other due to the dierent exponents, but y appears in both equations with the same exponent, 2. Thus, we should eliminate y 2 by multiplying the rst equation by 2 2 : 2x 2 + 2y x 2y 2 2x 2 2x 40. This looks like a quadratic equation. Let us move everything to one side and use the quadratic formula: 2x 2 2x 40 0 x ( 2) ± ( 2) 2 4(2)( 40) 2(2) 2 ± ± ± 8, 4 x , or x Let us nd the y-values for solutions that have x 5. Plug x 5 into an original equation, such as the rst one: (5) 2 + y y 2 7 y 2 8 This equation has no solutions. This means that the system does not actually have solutions where x 5. Let us nd the y-values for solutions that have x 4. Plug x 4 into an original equation, such as the rst one: ( 4) 2 + y y 2 7 y 2. 4

5 This equation has two solutions: y, or y. We got two y-values when plugging in x 4. This means that there are two solutions to the system that have x 4: x 4, y, x 4, y. Let us check both solutions. Plug x 4, y into both original equations: ( 4) 2 + () ( 4) 2() 2 8 2() 8 2 Hence x 4, y is a solution to the system. Plug x 4, y into both original equations: ( 4) 2 + ( ) ( 4) 2( ) 2 Hence x 4, y is a solution to the system. 8 2() 8 2 5

6 3. A System with Denominators Let us use the elimination method to solve x + 3 y 4 x + y 3 This is a system where x and y are in denominators (later when we learn about exponents we will see that x and y are being raised to negative powers). Let us move the numerators down in front of each fraction so they look like coecients: x + 3 y 4 x y 3 We may choose to eliminate either /x, or /y. Let us eliminate /y by multiplying the top equation by 3 2 : 2 x + y 4 x y x This is a fractional equation. We need to multiply both sides by x to get it out of the denominator: Solve for x by dividing by : 2 x x x 2 x 2 x. Plug x 2 into an original equation, such as the rst one: y Subtract /2 from both sides: 3 y

7 Multiply both sides by y: 3 y y 2 y Solve for y by dividing by /2, or multiplying by 2: 3 2 y. y. The solution to the system is x 2, y. Since we multiplied by variables along the way, we must check our answers by plugging them into both original equations: Hence x 2, y is a solution to the system

8 4. An Inconsistent System Let us solve 2x 3y 4 4x + 2y 9 Let us eliminate x by multiplying the rst equation by : 4x 2y 28 4x + 2y 9 0x + 0y The system has no solutions. If you try to use the elimination method on an inconsistent system of linear equations, then after adding the equations in Step 2, you get an equation where the variables are gone, and you have dierent numbers on each side. Indeed, notice that the graphs of the equations are parallel lines with the same slope but dierent y-intercepts. We can solve each equation for y to put them in slope-intercept form: 2x 3y 4 The rst line has slope 2/3 and y-intercept 4/3. 3y 2x + 4 y 2 3 x x + 2y 9 2y 4x + 9 y 4 2 x y 2 3 x The second line has slope 2/3 and y-intercept 3/7. Hence the lines are parallel and never intersect, so the system of equations has no solution. 8

9 5. A Dependent System Let us solve 2x 3y 4 4x + 2y 28 Let us eliminate x by multiplying the rst equation by : 4x 2y 28 4x + 2y 28 0x + 0y This equation is always true. If you try to use the elimination method on a dependent system of linear equations, then after adding the equations in Step 2, you get an equation where the variables are gone, and you have the same numbers (usually zero) on each side. Indeed, notice that the graphs of the equations are lines with the same slope and y-intercepts. We can solve each equation for y to put them in slope-intercept form: 2x 3y 4 The rst line has slope 2/3 and y-intercept 4/3. 3y 2x + 4 y 2 3 x x + 2y 28 2y 4x 28 y 4 2 x y 2 3 x 4 3. The second line has slope 2/3 and y-intercept 4/3. Hence the lines coincide, intersecting innitely many times along themselves, so the system of equations has innitely many solutions. 9

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