Simultaneous Linear Equations [PATCHED]

The coefficients of x in the two equations are 2 and 3 respectively. Let us multiply the first equation by 3 and the second equation by 2, so that the coefficients of x in the two equations become equal:

The point of intersection is $A\left( {2,\,\,2} \right)$, which means that $x = 2,\;\;y = 2$ is a solution to the pair of linear equations given by (2). In fact, it is the only solution to the pair, as two non-parallel lines cannot intersect in more than one point.

Simultaneous Linear Equations

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Two or more linear equations that all contain the same unknown variables are called a system of simultaneous linear equations. Solving such a system means finding values for the unknown variables which satisfy all the equations at the same time.

There are two common methods for solving simultaneous linear equations: substitution and elimination. In some questions, one method is the more obvious choice, often because it makes the process of solving the equations simpler; in others, the choice of method is up to personal preference. In either case, both methods would eventually lead to the same solution.

1) Intersect there is one unique solution to the system of equations. This makes sense as there can only pair of values (one for each variable) which satisfy both equations at the same time. This pair of value will correspond to the point of intersection.

2) Are the same line there are said to be infinitely many solutions to the system of equations: all pairs of values for the variables on one line satisfy the equation of the other line as they are the same line.

3) Are parallel there are no solutions: there are no pairs of values of $x$ and $y$ which lie on both lines (as the lines can never cross!) and so there can be no pairs of values for the variables which satisfy both equations.

One way to determine whether a system of two linear equations can be solved and how many solutions it has is to arrange both equations into the form: \[y=ax+b\] where $a$ is the slope of the line and $b$ is the $y$-axis intercept. It should now be clear whether the two equations correspond to either:

The substitution method involves substituting one equation into another in order to eliminate one of the variables. We can then solve for the other variable. The next step is to substitute the value of this variable into one of the equations to determine the value of the other variable. Note that this may involve rearranging one of the equations so that is it in a form which can easily be substituted into the other equation.

where equation $\textbf{(3)}$ corresponds to equation $\textbf{(1)}$ and equation $\textbf{(4)}$ corresponds to equation $\textbf{(2)}$. We can now see that the two equations correspond to different and non-parallel lines (as they have different slopes) and so there is one unique solution to this system of equations. We will now solve this system using the substitution method.

This graph shows the lines $y-2x=1$ and $2y-3x=5$. The solution to these simultaneous equations is the point at which the two lines intersect. We can see that this is the point $(3,7)$, which agrees with our solution.

The remaining equation is in one variable, and can be solved in the usual way; the value of that variable is then substituted into one of the original equations to find the value of the eliminated variable.

This graph shows the lines $2x+y=7$ and $3x-y=8$. The solution to these simultaneous equations is the point at which the two lines intersect. We can see that this is the point (3,1) which agrees with our solution.

The above methods for solving pairs of simultaneous linear equations can used to find the market equilibrium when we have linear market supply and market demand functions. For example, suppose that in a small economy the market demand and supply functions for bananas are: \begin{align} &q^S=5p-25\\ &q^D=-2p+24 \end{align} where $q^D$ is the quantity of bananas demanded (in kilos), $q^S$ is the quantity of bananas supplied (in kilos) and $p$ is the price per kilo of bananas (in $$).

This market is in equilibrium when the quantity demanded is equal to the quantity supplied: \[q^D=q^S\] As here we are only interested in this market when it is in equilibrium, in order to solve the system of equations we can set both $q^D$ and $q^S$ equal to $q$. The demand and supply functions then become: \begin{align} &q=5p-25 & \textbf{(1)}\\ &q=-2p+24 & \textbf{(2)} \end{align} To find the equilibrium, we can use either of the above methods to find the values of the variables $q$ and $p$ which satisfy both of these functions. Here we will use the substitution method.

So the solution to this system of equations is $q=10$ and $p=7$. This solution is shown graphically below, where we have plotted the inverse demand function (in blue) and the inverse supply function (in red) on the same graph. We can see that the equilibrium corresponds to the intersection of these two functions. That is, the point where $q=10$ and $p=7$.

Another way to express this information is in the form of equations: \begin{align} x^*&=x-0.18x-0.03y\\ y^*&=y-0.21x-0.04y\\ \end{align} where $x$ and $y$ are the total amounts (in kg) or gross output of coal and steel produced and $x^*$ and $y^*$ are the net amounts (in kg) of steel and coal produced.

Now, suppose that to satisfy the consumption needs of this economy, $36.06$kg of steel and $26.53$kg of coal must be produced. This is the net output of steel and coal respectively and so we can rewrite (and label) the above equations as: \begin{align} 26.53&=0.82x-0.03y & \textbf{(1)} \\ 36.06&=0.96y-0.21x& \textbf{(2)} \\ \end{align} We now have a system of simultaneous linear equations which we can solve using either of the above methods. As usual, the first step is to rewrite the equations in the form $y=mx+c$. This gives: \begin{align} y&=\dfrac{82x}{3}-\dfrac{26.53}{0.03} & \textbf{(3)}\\ y&=\dfrac{7x}{32}+\frac{36.06}{0.96} & \textbf{(4)}\\ \end{align}

Here we will look at how methods for solving systems of simultaneous equations can be used to determine the equilibrium level of income for the whole economy. To do this, we must first consider the key relationships in our model of the economy.

If we make the simplifying assumption that households earn their income based solely upon how much output they produce, then aggregate household income $Y$ must equal output $Q$. We can write this as an identity \[Y\equiv Q\] If we also assume that all output is purchased by someone we have that output $Q$ is equal to demand or aggregate expenditure $E$, so: \[Q\equiv E\] Combining the above equations gives: \[Y\equiv E\;\;\;\textbf{(1)}\] Now, since aggregate expenditure is made up of consumption expenditure by households $C$ and investment expenditure by firms $I$, we have another identity: \[E\equiv C+I\;\;\;\textbf{(2)}\]

The consumption function says that planned household consumption $\hat{C}$ is a function of household income. If we assume that this relationship is linear, then we can write: \[\hat{C}=aY+b\;\;\;\textbf{(3)}\] where $a$ and $b$ are both positive constants and $a\lt 1$.

a) As for the general model, we can simplify this model of the economy by combining equations to reducing the number of them. Combining the second equation with the first gives: \[Y\equiv C+250\;\;\;\textbf{(1)}\] where $250$ has been substituted for $I$. Combining the third equation with the fifth gives: \[C=0.25Y+50\;\;\;\textbf{(2)}\] We now have $2$ equations and $2$ unknowns so we can solve the system of equations. Substituting equation $\textbf{(2)}$ into equation $\textbf{(1)}$ gives: \begin{align} Y&=(0.25Y+50)+250\\ &=0.25Y+300\\ \Rightarrow 0.75Y&=300\\ \Rightarrow Y&=400 \end{align} and substituting this value of $Y$ into equation $\textbf{(2)}$ gives: \begin{align} C&=0.25\times 400+50\\ &=150 \end{align} so the equilibrium level of income is $Y=400$ and the equilibrium level of consumption is $C=150$.

b) If investment rises to $400$, equation $\textbf{(1)}$ from part a) becomes: \[Y\equiv C+550\;\;\;\textbf{(1.1)}\] and equation $\textbf{(2)}$ remains unchanged as it does not contain an $I$ term: \[C=0.25Y+50\;\;\;\textbf{(2.1)}\] We will now use the substitution method to solve these equations. Substituting equation $\textbf{(2)}$ into equation $\textbf{(1)}$ gives: \begin{align} Y&=(0.25Y+50)+550\\ &=0.25Y+600\\ \Rightarrow 0.75Y&=600\\ \Rightarrow Y&=450 \end{align} and substituting this value of $Y$ into equation $\textbf{(2.1)}$ gives: \begin{align} C&=0.25\times 450+50\\ &=162.50 \end{align} so when investment rises to $550$ the equilibrium level of income rises to $Y=450$ and the equilibrium level of consumption rises to $C=162.50$. This makes sense as investment is part of aggregate demand, so when investment rises output and therefore aggregate income should also rise.

In Mathematics, linear equations are the equations in which the highest power of the variable is one. The result of the linear equation is always a straight line. Thus, simultaneous linear equations are the system of two linear equations in two or three variables that are solved together to find a common solution. The simultaneous linear equation is represented as follows:

Here, a1 and a2 are the coefficients of x, and b1 and b2 are the coefficients of y, and c1 and c2 are constant. The solution for the system of linear equations is the ordered pair (x, y), which satisfies the given equation.

And we can find the matching value of y using either of the two original equations (because we know they have the same value at x=1). Let's use the first one (you can try the second one yourself):

Linear systems are the basis and a fundamental part of linear algebra, a subject used in most modern mathematics. Computational algorithms for finding the solutions are an important part of numerical linear algebra, and play a prominent role in engineering, physics, chemistry, computer science, and economics. A system of non-linear equations can often be approximated by a linear system (see linearization), a helpful technique when making a mathematical model or computer simulation of a relatively complex system. 17dc91bb1f