Question

1. \

$$\begin{cases} -2x + 5y = 7 \\\\ 2x - 5y = -7 \\end{cases}$$

1. \

$$\begin{cases} -10x + 2y = -7 \\\\ -10x + 2y = 0 \\end{cases}$$

Explanation:

Problem 12: Solve the system of equations

$$\begin{cases}-2x + 5y = 7\\2x - 5y = -7\end{cases}$$

Step 1: Add the two equations

Add the left - hand sides and the right - hand sides of the two equations together.
$(-2x + 5y)+(2x - 5y)=7+(-7)$
Simplify the left - hand side: $-2x+2x + 5y-5y=0$, and the right - hand side: $7 - 7 = 0$. So we get $0 = 0$.

Step 2: Analyze the result

When we add the two equations, we get a true statement ($0 = 0$) which means that the two equations are equivalent (they represent the same line). So the system of equations has infinitely many solutions. The general solution can be expressed by solving one of the equations for one variable in terms of the other. Let's solve the first equation for $x$:
From $-2x+5y = 7$, we have $-2x=7 - 5y$, then $x=\frac{5y - 7}{2}$, where $y$ can be any real number. Or we can solve for $y$: from $-2x + 5y=7$, we get $5y=2x + 7$, so $y=\frac{2x + 7}{5}$, where $x$ can be any real number.

Step 1: Subtract the two equations

Subtract the second equation from the first equation (or vice - versa). Let's subtract the second equation from the first:
$(-10x + 2y)-(-10x + 2y)=-7-0$

Step 2: Analyze the result

Simplify the left - hand side: $-10x + 2y + 10x-2y = 0$, and the right - hand side: $-7$. So we get $0=-7$, which is a false statement.

Answer:

The system has infinitely many solutions. The solutions are given by $x=\frac{5y - 7}{2}$ (or $y=\frac{2x + 7}{5}$) for all real numbers $x$ and $y$.

Problem 15: Solve the system of equations

$$\begin{cases}-10x + 2y=-7\\-10x + 2y = 0\end{cases}$$