Question

1. india wants to solve the system of linear equations shown, by elimination.\

$$\begin{cases}-3x + 2y = 16\\\\2x + 5y = 21\\end{cases}$$

which is her best first step?\bigcirc multiply the first equation by 3.\bigcirc multiply the second equation by 2.\bigcirc multiply the first equation by 3 and the second equation by 2.\bigcirc multiply the first equation by 2 and the second equation by 3.

Explanation:

Step1: Recall Elimination Method

The elimination method aims to eliminate one variable by making the coefficients of that variable equal (in absolute value) in both equations. Here, we look at the coefficients of \(x\): \(-3\) (in first equation) and \(2\) (in second equation).

Step2: Analyze Coefficient Adjustment

To eliminate \(x\), we need to make the coefficients of \(x\) such that when we add or subtract the equations, \(x\) terms cancel. The least common multiple of \(3\) and \(2\) (the absolute values of coefficients of \(x\)) is \(6\). So, we multiply the first equation by \(2\) (to get \(-6x\)) and the second equation by \(3\) (to get \(6x\)). Then, when we add the two new equations, the \(x\) terms will cancel (\(-6x + 6x = 0\)).

Let's check each option:

• Option 1: Multiply first equation by 3. New first equation: \(-9x + 6y = 48\). Coefficient of \(x\) is \(-9\), second equation has \(2x\). Not helpful for elimination.
• Option 2: Multiply second equation by 2. New second equation: \(4x + 10y = 42\). First equation has \(-3x\). Not helpful.
• Option 3: Multiply first by 3 (\(-9x + 6y = 48\)) and second by 2 (\(4x + 10y = 42\)). Coefficients of \(x\) are \(-9\) and \(4\), not canceling.
• Option 4: Multiply first by 2: \(2(-3x + 2y)=2(16)\Rightarrow -6x + 4y = 32\). Multiply second by 3: \(3(2x + 5y)=3(21)\Rightarrow 6x + 15y = 63\). Now, adding these two equations: \((-6x + 6x)+(4y + 15y)=32 + 63\), which eliminates \(x\).

Answer:

D. Multiply the first equation by 2 and the second equation by 3.