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In this video, we will learn how to solve linear-quadratic systems of equations.
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Letβs begin by discussing what a linear-quadratic system of equations actually is.
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Itβs not as scary as it sounds.
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Itβs a system of two equations in which one is a linear equation and the other is a quadratic.
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Remember, a linear equation is one in which the highest power of each variable that appears is one and none of the variables are multiplied together.
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For example, the equation π¦ equals two π₯ is a linear equation.
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And if you were to draw its graph, it would be represented by a straight line.
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A quadratic equation, however, is one in which there will be a least one squared term.
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For example, the equation π₯ squared plus π¦ squared equals five is a quadratic equation.
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We may also see equations which include terms when the two variables are multiplied together, for example, the equation π₯ plus two π₯π¦ equals three.
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If you were to plot the graph of a quadratic equation, it gives some kind of curve.
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In the case of π₯ squared plus π¦ squared equals five, it is a circle.
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From a graphical point of view, solving a linear-quadratic system of equations is equivalent to finding the coordinates of any points of intersection between the two graphs.
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Now, weβre going to mostly be using the method of solving simultaneous equations by substitution.
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So you should make sure that youβre already familiar and comfortable with this method.
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You should also be comfortable solving quadratic equations in one variable by factoring.
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Weβll look at some applications of these techniques to worded problems and problems involving points of intersection of straight lines and curves.
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Letβs consider our first example then.
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Solve the simultaneous equations π¦ equals π₯ minus two, π₯ minus two squared plus π¦ minus three squared is equal to nine.
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The first thing to notice is that this is a linear-quadratic system of equations.
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The first equation π¦ equals π₯ minus two is a linear equation, as the highest power of π₯ and π¦ that appears is one.
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And the second equation π₯ minus two squared plus π¦ minus three squared equals nine is a quadratic equation because once weβve distributed the parentheses, we will have both π₯ squared and π¦ squared terms.
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Weβre going to use the method of substitution to answer this problem.
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Now, our first equation is π¦ equals π₯ minus two.
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And we notice that the expression π₯ minus two appears in the second equation.
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So what we can do is replace π₯ minus two with π¦ in our second equation.
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Doing so gives π¦ squared plus π¦ minus three squared equals nine.
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And so we have an equation in π¦ only.
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Itβs a quadratic equation which we can solve.
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We begin by distributing the parentheses in π¦ minus three squared.
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And we remember that π¦ minus three all squared means π¦ minus three multiplied by π¦ minus three.
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So when we distribute the parentheses, we have four terms.
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And these four terms then simplify to π¦ squared minus six π¦ plus nine.
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Collecting like terms on the left-hand side, we have the quadratic equation two π¦ squared minus six π¦ plus nine equals nine.
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Now, we notice that as we have positive nine on each side, these two terms will directly cancel one another out, which is equivalent to subtracting nine from each side of the equation.
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This leaves us with the simplified equation two π¦ squared minus six π¦ equals zero, which we can solve by factoring.
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The highest common factor of two π¦ squared and negative six π¦ is two π¦.
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And to make two π¦ squared, we have to multiply two π¦ by π¦.
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And to make negative six π¦, we have to multiply two π¦ by negative three.
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So our quadratic in its factored form is two π¦ multiplied by π¦ minus three is equal to zero.
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To solve, we take each factor in turn, set it equal to zero, and then solve the resulting linear equation.
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The first equation is two π¦ equals zero, which we can solve by dividing each side by two to find that π¦ is equal to zero.
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The second equation is π¦ minus three equals zero.
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We solve by adding three to each side giving π¦ equals three.
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So we find that our solution to these simultaneous equations includes two π¦-values, π¦ equals zero and π¦ equals three.
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We now need to find the corresponding π₯-values.
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And to do this, we substitute each π¦-value into the linear equation.
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It must be the linear equation into which we substitute because itβs possible that, on the quadratic curve, there is more than one point where π¦ equals zero or π¦ equals three.
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But on the linear graph, the straight line, there will only be one point where π¦ equals zero and one point where π¦ equals three.
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So we must substitute into the linear equation.
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When π¦ equals zero, we have zero equals π₯ minus two.
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And adding two to each side, we find that π₯ is equal to two.
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When π¦ equals three, we have that three is equal to π₯ minus two.
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And adding two to each side, we find that π₯ is equal to five.
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So the solution to this pair of simultaneous equations is two pairs of π₯π¦-values.
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π₯ equals two and π¦ equals zero and π₯ equals five and π¦ equals three.
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Itβs important to understand that these solutions come in pairs.
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And we canβt mix and match the π₯- and π¦-values.
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For example, π₯ equals two and π¦ equals three is not a valid solution to this pair of simultaneous equations, which you can confirm by substituting the values into the linear equation or, indeed, the quadratic.
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Remember, we used the method of substitution to answer this question by replacing π₯ minus two with π¦.
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It would also have been possible to replace π¦ in the second set of parentheses with π₯ minus two to give an equation in π₯ only.
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In this case, weβd have found our solutions for π₯ first and then substitute it back into the linear equation to find our solutions for π¦.
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Our final answer is that π₯ equals two and π¦ equals zero or π₯ equals five and π¦ equals three.
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In our next example, weβll see how we can apply these methods to a worded problem.
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The sum of two numbers is 11 and the sum of their squares is 65.
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What are the numbers?
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Now, itβs important to notice that we shouldnβt answer this question using trial and error.
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We need to use a formal approach.
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Weβre going to use some algebra to answer the problem.
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So weβll let the two numbers be represented by the letters π₯ and π¦.
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Weβre now going to express the information in the question as equations involving π₯ and π¦.
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Firstly, the sum of the two numbers is 11.
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So this gives us the equation π₯ plus π¦ is equal to 11.
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Secondly, weβre told that the sum of their squares is 65.
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So this gives us the equation π₯ squared plus π¦ squared equals 65.
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We now have a linear-quadratic system of equations.
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The first equation is linear, and the second is quadratic.
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Weβre going to use the method of substitution to solve these two equations simultaneously.
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We begin by rearranging the linear equation to give one variable in terms of the other.
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And in this problem, it doesnβt make any difference which variable we choose because the problem is equally complicated or equally simple in both variables.
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Iβve chosen to rewrite equation one as π¦ equals 11 minus π₯.
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We now take this expression for π¦ in terms of π₯ and substitute it into the second equation, thatβs into the quadratic equation.
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Giving π₯ squared plus 11 minus π₯ all squared is equal to 65.
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And now, we have a quadratic equation in one variable π₯ only.
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We can distribute the parentheses carefully, remembering that 11 minus π₯ all squared means 11 minus π₯ multiplied by 11 minus π₯, which simplifies to 121 minus 22π₯ plus π₯ squared.
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We can group the like terms on the left-hand side β thatβs π₯ squared plus π₯ squared β giving two π₯ squared.
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And at the same time, subtract 65 from each side to give the quadratic equation two π₯ squared minus 22π₯ plus 56 is equal to zero.
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Now, we notice at this point that each of the coefficients in our equation are even numbers.
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So we can simplify by dividing the entire equation by two.
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Doing so gives the simplified quadratic equation π₯ squared minus 11π₯ plus 28 is equal to zero.
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Now, we want to solve this equation for π₯.
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So we first look to see whether this equation can be factored.
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As the coefficient of π₯ squared in our equation is one, the first term in each set of parentheses will simply be π₯ because π₯ multiplied by π₯ gives π₯ squared.
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And to complete the parentheses, weβre then looking for two numbers whose sum is the coefficient of π₯ β thatβs negative 11 β and whose product is the constant term β thatβs 28.
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By considering the factors of 28, we see that if we choose the two numbers negative seven and negative four, then their product is indeed 28 because a negative multiplied by a negative gives a positive.
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And their sum is indeed negative 11.
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So these are the two numbers weβre looking for to complete our parentheses.
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We have our quadratic in its factored form then.
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π₯ minus seven multiplied by π₯ minus four is equal to zero.
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To solve, we take each factor in turn, set it equal to zero, and solve the resulting equation.
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We have π₯ minus seven equals zero, which can be solved by adding seven to each side to give π₯ equals seven, and then π₯ minus four equals zero, which we solve by adding four to each side to give π₯ equals four.
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We found then the solution for π₯.
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There are two possible values.
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π₯ is either equal to four or π₯ is equal to seven.
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We now need to find the corresponding π¦-values, which we can do by substituting each π₯-value in turn into the linear equation which, remember, that was π¦ is equal to 11 minus π₯.
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When π₯ is equal to seven, we find that π¦ is equal to 11 minus seven which is four.
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And when π₯ is equal to four, we find that π¦ is equal to 11 minus four, which is seven.
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So we have exactly the same values for π¦, as we did for π₯.
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The reason for this is because at the beginning of the problem, we just let the two numbers be represented by π₯ and π¦.
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We didnβt specify whether π₯ or π¦ was the larger number.
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So we found the same solution twice.
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The two numbers are seven and four.
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And either can be π₯ and then the other will be π¦.
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We can of course check our answer, firstly, by confirming that the sum of our two numbers is 11, which of course it is, and, secondly, by confirming that the sum of their squares β seven squared plus four squared, which is 49 plus 16 β is indeed equal to 65.
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So by first formulating the information in the question as a linear-quadratic system of equations and then solving this pair of simultaneous equations using the substitution method, we found that the two numbers weβre looking for are seven and four.
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Now, one really useful application of these techniques is in finding the points of intersection of a straight line and a curve, as mentioned right at the start of the video.
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Letβs now consider an example of this type.
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Find the set of points of intersection of the graphs of π₯ minus π¦ equals zero and six π₯ squared minus π¦ squared equals 45.
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So weβve been asked to find the points of intersection of π₯ minus π¦ equals zero, which is a straight line, and six π₯ squared minus π¦ squared equals 45, which is a curve.
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This is equivalent to solving the linear-quadratic system of equations π₯ minus π¦ equals zero, six π₯ squared minus π¦ squared equals 45.
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Weβre going to do this using the method of substitution.
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We begin by rearranging the linear equation to give one variable in terms of the other.
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We find that π₯ is equal to π¦.
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Weβre now going to substitute our expression for π₯ into the second equation.
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Doing so gives six π¦ squared minus π¦ squared is equal to 45.
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We could equally have substituted π¦ equals π₯ into our second equation, which would give six π₯ squared minus π₯ squared equals 45.
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Both approaches will lead us to the same solution.
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We can now solve this quadratic equation for π¦.
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Simplifying the left-hand side, six π¦ squared minus π¦ squared, gives five π¦ squared.
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We can then divide through by five, giving π¦ squared equals nine and solve by square rooting.
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Remembering, we must take plus or minus the square root.
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So we have that π¦ is equal to plus or minus the square root of nine, which is positive or negative three.
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Having found our π¦-values, we now need to find the corresponding π₯-values by substituting into the linear of equation.
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And itβs very straightforward.
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As our linear equation can be expressed as π₯ equals π¦, then each π₯-value is just the same as the corresponding π¦-value.
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We find then that there are two points of intersection between these graphs, the point three, three and the point negative three, negative three, which we can express as the set containing these two coordinates.
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Now, you may not immediately recognize what the graph of six π₯ squared minus π¦ squared equals 45 looks like.
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But if you have access to a graphics calculator or some graphical plotting software, then you can plot these two graphs.
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π₯ minus π¦ equals zero is a straight line and six π₯ squared minus π¦ squared equals 45 is whatβs known as a hyperbola.
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And by considering these two graphs, you can confirm that the two points of intersection are indeed the points weβve given here.
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In our final example, weβll consider a problem where the two variables are multiplied together in one of the two equations.
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Given that π₯ squared plus π₯π¦ equals 18 and π₯ plus π¦ equals six, find the value of π₯.
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What we have here is a linear-quadratic system of equations.
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The first equation π₯ plus π¦ equals six is linear.
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And the second equation π₯ squared plus π₯π¦ equals 18 is quadratic because it includes an π₯ squared term and also the term π₯π¦, where the two variables are multiplied together.
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Weβre not asked to fully solve this system of equations but simply to find the value of π₯.
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So weβre going to do this using the substitution method.
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We begin by rearranging the first equation to give π¦ equals six minus π₯ because this gives an expression for π¦ in terms of π₯, which we can substitute into the second equation to give an equation in π₯ only.
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Doing so gives the equation π₯ squared plus π₯ multiplied by six minus π₯ equals 18.
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And we now have a quadratic equation in π₯, which we can solve.
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We distribute the parentheses on the left-hand side to give π₯ squared plus six π₯ minus π₯ squared equals 18.
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And we now see that the π₯ squared and negative π₯ squared terms will cancel each other out.
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So, in fact, our equation reduces to a linear equation.
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We have six π₯ is equal to 18.
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And this equation can be solved by dividing each side by six to give π₯ equals three.
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So by substituting π¦ equals six minus π₯ into the second equation, we created an equation in π₯ only which we could then solve to find the value of π₯.
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Of course, we arenβt asked to find the value of π¦ in this problem.
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But if we did need to, we could substitute the value of π₯ weβve just found back into our linear equation, π¦ equals six minus π₯, to find the corresponding value of π¦.
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Our solution to the problem is that the value of π₯ is three.
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Letβs now summarize what weβve seen in this video.
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Firstly, a linear-quadratic system of equations is simply a system of two equations in which one is linear and the other is quadratic.
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We solve such systems of equations by using the method of substitution.
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We rearrange the linear equation to give one variable in terms of the other, and then we substitute our expression for this variable into the quadratic equation.
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We then solve the resulting quadratic equation, usually by factoring.
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We then substitute the value or values we found for the first variable back into the linear equation to find the corresponding values for the second.
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We must remember that our solutions come in pairs.
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So we must give our solution as corresponding pairs of the two variables.
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We canβt mix and match the different values.
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We saw also in this video that this method can be applied to number problems.
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And a particularly useful application is in finding the coordinates of any points of intersection between a straight line and a curve.