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Simultaneous equations are a set of equations with multiple variables, which are solved together to find a solution for each variable. Here, we had two equations involving the variables \(x\) and \(y\):

• Equation (1): \(5x + 7y = 10\)
• Equation (2): \(3x - 14y = 6\)

The idea is to solve both equations at the same time, hence the name 'simultaneous'. By solving them together, we find values for \(x\) and \(y\) that fit both equations. This type of problem is common in algebra and can be solved through various methods, one of which is the **addition method** used in this exercise. The addition method helps eliminate one variable by creating opposite coefficients. After eliminating one variable, we can easily solve for the other.

Linear equations like the ones in our exercise are equations of the first order. This means each term is either a constant or the product of a constant and a single variable. In the context of our simultaneous equations, solving a linear equation involves finding the values of variables that make both equations true simultaneously.The linear form is expressed generally as \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. The solutions to these equations describe lines on a graph, and the intersection of these lines represents the solution to the equations—a point where both equations are satisfied.

In our example, we used linear equation solving by manipulating the equations to eliminate one variable and solve for the other, then substituting back to find the complete solution.

Algebraic manipulation is a technique used to rearrange and simplify equations. This is essential in solving simultaneous equations. In solving the given equations using the addition method, algebraic manipulation was crucial.

• **Step 1:** Both equations were manipulated by multiplying them to create equal and opposite coefficients for \(y\). We multiplied the entire first equation by 2, resulting in \(10x + 14y = 20\), and the second equation by 5, obtaining \(15x - 70y = 30\).
• **Step 2:** After manipulation, we added these equations together to eliminate \(y\), simplifying our system to a single equation \(25x = 50\).
• **Step 3:** With algebraic simplification, we immediately found \(x = 2\). Substituting back, we determined \(y = 0\).

These steps of multiplying, adding, and substituting exemplify algebraic manipulation. Such techniques are fundamental skills in algebra and are often needed to solve complex equation sets efficiently. By practicing these manipulations, students build a solid foundation for tackling more advanced mathematical problems.