Multiply the first equation by 4 and the second by 3 to eliminate $ y $:
["How to Eliminate y by Scaling Equations: A Step-by-Step Guide", "Understanding how to eliminate variables is a crucial skill in solving systems of equations, especially in algebra and precalculus. One effective method is multiplying each equation by a suitable factor so that when you add or subtract both equations, one variable cancels out. In this article, we’ll explore how multiplying the first equation by 4 and the second by 3 helps eliminate the variable ( y ), making it easier to solve for ( x ) and ( y ).", "## Why Eliminate Behind Systems of Equations?", "When solving systems of linear equations, eliminating a variable simplifies the problem from two equations in two unknowns to one equation with one unknown, significantly reducing complexity. Elimination by scaling is especially powerful when direct substitution is cumbersome.", "### The Basic Problem", "Suppose we have two equations:", "[\n\begin{align}\na_1x + b_1y &= c_1 \quad \ ext{(Equation 1)} \\na_2x + b_2y &= c_2 \quad \ ext{(Equation 2)}\n\end{align}\n]", "If the goal is to eliminate ( y ), the coefficients of ( y ) must become opposites. Since the original coefficients are ( b_1 ) and ( b_2 ), multiplying each equation by appropriate constants aligns these so that when added or subtracted, ( y ) vanishes.", "### Step 1: Multiply First Equation by 4", "Multiply the first equation by 4:", "[\n4(a_1x + b_1y) = 4c_1 \quad \Rightarrow \quad 4a_1x + 4b_1y = 4c_1\n]", "Now, the second equation remains unchanged for now:", "[\na_2x + b_2y = c_2\n]", "### Step 2: Multiply Second Equation by 3", "This step is critical — multiplying by 3 ensures the ( y )-coefficients become opposites:", "[\n3(a_2x + b_2y) = 3c_2 \quad \Rightarrow \quad 3a_2x + 3b_2y = 3c_2\n]", "Now we have:", "[\n\begin{align}\n4a_1x + 4b_1y &= 4c_1 \quad \ ext{(8x + My = N)} \\n3a_2x + 3b_2y &= 3c_2 \quad \ ext{(Px + Qy = R)}\n\end{align}\n]", "Since ( 4b_1 ) and ( 3b_2 ) are negative opposites (i.e., ( 4b_1 = -3b_2 )), subtracting the equations eliminates ( y ):", "[\n(4a_1x - 3a_2x) + (4b_1y - 3b_2y) = 4c_1 - 3c_2\n]", "[\n(4a_1 - 3a_2)x + 0 = 4c_1 - 3c_2\n]", "### Step 3: Solve for ( x )", "Now solve:", "[\n(4a_1 - 3a_2)x = 4c_1 - 3c_2\n]", "[\nx = \frac{4c_1 - 3c_2}{4a_1 - 3a_2}\n]", "### Step 4: Back-Substitute to Find ( y )", "Once ( x ) is found, substitute back into either original equation. Using Equation 1:", "[\n4a_1\left( \frac{4c_1 - 3c_2}{4a_1 - 3a_2} \right) + 4b_1y = 4c_1\n]", "Solve for ( y ) to complete the solution.", "---", "### Summary", "Multiplying the first equation by 4 and the second by 3 is a strategic way to scale the equations so that the coefficients of ( y ) are opposites. This allows elimination, reducing the system to a single equation in ( x ), making it faster and simpler to solve.", "### Key Takeaways", "- Choose scaling factors so that the coefficients of the variable to eliminate become negatives of each other.\n- Multiplying equations by constants preserves equality and does not change the solution set.\n- Elimination via scaling is efficient when direct substitution is complex or inefficient.", "Mastering this technique strengthens your ability to tackle systems of equations quickly and accurately — a valuable skill in math, engineering, and applied sciences.", "---", "Keywords: eliminate y, eliminate variable, solving systems of equations, algebra technique, linear equations, scaling equations, matrix methods, elimination method, math tutorial, precalculus", "Meta Description: Learn how multiplying the first equation by 4 and the second by 3 eliminates ( y ) in systems of equations—step-by-step guide for students and math enthusiasts. Improve your solving skills with this powerful elimination technique."]
