Solving Problems By Elimination

Felix may notice that now both equations have a constant of 25, but subtracting one from another is not an efficient way of solving this problem.

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The elimination method of solving systems of equations is also called the addition method.

To solve a system of equations by elimination we transform the system such that one variable "cancels out".

So let’s now use the multiplication property of equality first.

You can multiply both sides of one of the equations by a number that will result in the coefficient of one of the variables being the opposite of the same variable in the other equation. Notice that the first equation contains the term 4y, and the second equation contains the term y.

Example 3: $$ \begin 2x - 5y &= 11 \\ 3x 2y &= 7 \end $$ Solution: In this example, we will multiply the first row by -3 and the second row by 2; then we will add down as before.

$$ \begin &2x - 5y = 11 \color\ &\underline \end\ \begin &\underline} \text\ &19y = -19 \end $$ Now we can find: back into first equation: $$ \begin 2x - 5\color &= 11 \ 2x - 5\cdot\color &= 11\ 2x 5 &= 11\ \color &\color \color \end $$ The solution is $(x, y) = (3, -1)$.

Recall that a false statement means that there is no solution.

If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. A theater sold 800 tickets for Friday night’s performance. Combining equations is a powerful tool for solving a system of equations.

Substituting the value of y = 3 in equation (i), we get 2x 3y = 11 or, 2x 3 × 3 = 11or, 2x 9 = 11 or, 2x 9 – 9 = 11 – 9or, 2x = 11 – 9or, 2x = 2 or, x = 2/2 or, x = 1Therefore, x = 1 and y = 3 is the solution of the system of the given equations. Solve 2a – 3/b = 12 and 5a – 7/b = 1 Solution: The given equations are: 2a – 3/b = 12 ……………

(iv) Multiply equation (iii) by 5 and (iv) by 2, we get 10a – 15c = 60 …………… (vi) Subtracting (v) and (vi), we get or, c = 58 /-29 or, c = -2 But 1/b = c Therefore, 1/b = -2 or b = -1/2 Subtracting the value of c in equation (v), we get 10a – 15 × (-2) = 60 or, 10a 30 = 60 or, 10a 30 - 30= 60 - 30 or, 10a = 60 – 30 or, a = 30/10 or, a = 3 Therefore, a = 3 and b = 1/2 is the solution of the given system of equations. x/2 2/3 y = -1 and x – 1/3 y = 3 Solution: The given equations are: x/2 2/3 y = -1 …………… (ii) Multiply equation (i) by 6 and (ii) by 3, we get; 3x 4y = -6 …………… (iv) Solving (iii) and (iv), we get; or, y = -15/5 or, y = -3 Subtracting the value of y in (ii), we get; x - 1/3̶ × -3̶ = 3 or, x 1 = 3 or, x = 3 – 1 or, x = 2 Therefore, x = 2 and y = -3 is the solution of the equation.