Solution

It's a good idea to label the equations so they're easier to refer to. \begin{align} y\,-\,2x &= 1 & \textbf{(1)} \\ 2y-3x &=5 & \textbf{(2)} \end{align}

Rearrange equation 1 and make $y$ the subject.

\[y=1+2x\]

Substitute this expression for $y$ into equation 2.

\[2(1+2x)-3x=5\]

Now we have an equation in only one variable, which can be rearranged and solved to find $x$.

\begin{align} 2(1+2x)-3x&=5 \\ 2+4x-3x&=5\\ 2+x&=5\\ x&=3 \end{align}

Now we have a value for $x$, this can be substituted into either of the original equations to find $y$. It's easiest to use the rearranged expression for $y$ we derived from equation 1.

\[y= 1+2\times 3=7.\]

The solution $x=3$ and $y=7$ satisfies both equations. It's a good idea to check your solution by substituting these values of $x$ and $y$ into the original equations.

\begin{align} y - 2x &= 1 & \textbf{(1)}\\ 7 - 2 \times 3 &= 7 - 6 = 1. \\ \\ 2y - 3x &= 5 & \textbf{(2)}\\ 2 \times 7 - 3 \times 3 &= 14 - 9 = 5. \end{align}

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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.

Solution

Label the equations 1, 2 and 3, respectively.

\begin{array}{rcrcrcrcr} x + y - z = 4 && \textbf{(1)} \\ x - 2y + 3z = -6 && \textbf{(2)} \\ 2x + 3y + z = 7 && \textbf{(3)} \end{array}

Rearrange equation 1 to make $z$ the subject.

\[z=x+y-4\]

Substitute this expression for $z$ into equations 2 and 3.

Equation 2 becomes

\begin{align} x-2y+3(x+y-4)&=-6\\ x-2y+3x+3y-12&=-6\\ 4x+y&=6 & \textbf{(2*)} \end{align}

And equation 3 becomes

\begin{align} 2x+3y+(x+y-4)&=7\\ 2x+3y+x+y-4=7\\ 3x+4y&=11 & \textbf{(3*)} \end{align}

Now we only have two equations and two unknowns so we can solve as in the last example.

Rearrange equation 2* to make $y$ the subject.

\[y=6-4x\]

Substitute this into 3*

\begin{align} 3x+4(6-4x)&=11\\ 3x+24-16x&=11\\ -13x&=-13\\ x&=1 \end{align}

Now substitute this value for $x$ into 2* to give

\[y=6-4\times 1=2.\]

Now we know the values of two of the variables, we can substitute these both back into equation 1 to find $z$.

\[z=x+y-4=1+2-4=-1.\]

We have found that the solution to the system of simultaneous equations is $x=1$, $y=2$ and $z=-1$.

Solution

Notice here that the coefficient of $y$ is the same up to sign in both of the equations. As one is positive and one is negative, if we add the two equations together, the $y$'s will cancel.

\begin{align} &2x+y=7\\ &3x-y=8\\ &\overline{5x \quad\;\;\, = 15} \end{align}

So $x=3$.

Now substitute this value for $x$ back into one of the original equations.

\begin{align} 2x + y &= 7 \\ 2\times 3+y&=7\\ 6+y &=7\\ y&=1 \end{align}

The values $x=3$, $y=1$ satisfy both of the equations we were given.

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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.

Solution

Label the equations 1 and 2, respectively.

\begin{align} 2x + 3y &= 4 & \textbf{(1)} \\ x - 2y &= -5 & \textbf{(2)} \end{align}

We want to make the coefficients of $x$ the same in both equations. To do this, multiply equation (2) by $2$.

\[2x-4y=-10\]

As the coefficients of $x$ in both equations are positive, we subtract equation 2 from equation 1.

\begin{align} &2x+3y=4\\ &2x-4y=-10\\ &\overline{\qquad \; 7y =14} \end{align}

So $y=2$.

Substituting this value for $y$ back into one of the original equations will give $x$.

\begin{align} x - 2y &= -5 & \textbf{(2)} \\ x-2\times 2&=-5\\ x-4&=-5\\ x&=-1 \end{align}

The values $x=-1$ and $y=2$ satisfy both of the equations we were given.

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This graph shows the lines $2x+3y=4$ and $x-2y=-5$. The solution to these simultaneous equations is the point at which the two lines intersect.