Question

which value, when placed in the box, would result in a system of equations with infinitely many solutions? y = 2x - 5 2y - 4x = 0 0 -10 5 -5 10

Explanation:

Step1: Rewrite the first - type equation

The first equation is \(y - 2x=- 5\), and the second equation is \(y-4x = 0\). A system of linear equations \(a_1x + b_1y=c_1\) and \(a_2x + b_2y=c_2\) has infinitely many solutions when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Let the first equation be \(y - 2x=-5\) or \(-2x + y=-5\), and the second equation be \(-4x+ y = 0\). If we want the two - equations to be the same (for infinitely many solutions), we can rewrite the equations in the general form \(Ax+By = C\).

Step2: Make the two equations equivalent

Let's assume the second equation is \(ky-4x = 0\). We want it to be a multiple of \(y - 2x=-5\). If we multiply \(y - 2x=-5\) by \(2\), we get \(2y-4x=-10\). For the system \(

$$\begin{cases}y - 2x=-5\\ky-4x = 0\end{cases}$$

\), we can rewrite the first equation as \(2y-4x=-10\). For the two equations \(2y-4x=-10\) and \(ky - 4x=0\) to be the same (infinitely many solutions), we set \(k = 2\). But if we rewrite the equations in slope - intercept form \(y=mx + b\). The first equation is \(y=2x - 5\), and the second is \(y = 4x\). A system of linear equations \(y=m_1x + b_1\) and \(y=m_2x + b_2\) has infinitely many solutions when \(m_1=m_2\) and \(b_1 = b_2\). Let's rewrite the second equation as \(y=\frac{4}{k}x\) (from \(ky-4x = 0\)). From \(y = 2x-5\), we want \(\frac{4}{k}=2\).
Solve the equation \(\frac{4}{k}=2\) for \(k\). Cross - multiply: \(2k=4\), then \(k = 2\).

Answer:

2