Algebra - Simultaneous Linear Equations and their Applications

Solve the following system of equations using the elimination method: $3x + 2y = 12$ $5x - 2y = 4$

Step 1: Add the two equations to eliminate $y$: $\begin{array}{r} 3x + 2y = 12 \\ + (5x - 2y = 4) \\ \hline 8x = 16 \end{array}$ Step 2: Solve for $x$: $x = \frac{16}{8} = 2$ Step 3: Substitute $x = 2$ into the first equation to find $y$: $3(2) + 2y = 12$ $6 + 2y = 12$ $2y = 6$ $y = 3$ The solution is $(2, 3)$.

Since the coefficients of $y$ are opposites ($+2$ and $-2$), adding the equations cancels $y$ out immediately, allowing us to solve for $x$ first.

Solve using the substitution method: $y = 2x - 3$ $4x + 3y = 21$

Step 1: Substitute the expression for $y$ from the first equation into the second equation: $4x + 3(2x - 3) = 21$ Step 2: Expand and solve for $x$: $4x + 6x - 9 = 21$ $10x - 9 = 21$ $10x = 30$ $x = 3$ Step 3: Substitute $x = 3$ back into the first equation to find $y$: $y = 2(3) - 3$ $y = 6 - 3 = 3$ The solution is $(3, 3)$.

The first equation was already solved for $y$, making substitution the most efficient method to reduce the system to a single-variable equation.

The sum of two numbers is 20 and their difference is 4. Find the numbers.

Let the two numbers be $x$ and $y$. Based on the problem, we have: $x + y = 20$ $x - y = 4$ Adding the two equations: $\begin{array}{r} x + y = 20 \\ + (x - y = 4) \\ \hline 2x = 24 \end{array}$ $x = 12$ Substitute $x = 12$ into the first equation: $12 + y = 20$ $y = 8$ The two numbers are 12 and 8.

By translating the word problem into two algebraic equations, we can use the elimination method to find the values of the two unknown numbers.