1.8 Solving systems of linear equations and polynomial equations with technologyIB Maths: Applications and Interpretation HL: Revision notes

A system of linear equations is a set of equations that must all hold at the same time. In IB exams you solve them with technology and no method is prescribed. Systems have up to 3 variables and there is always a unique solution. First turn the context into equations. Example: three adults and two children pay 98 USD and two adults and five children pay 124 USD: $3a+2c=98,\qquad 2a+5c=124.$ Enter the equations in the GDC equation solver (or the simultaneous equations app, or a matrix) to get $a=22$, $c=16$. Then answer the question that was asked: four adults and three children cost $4(22)+3(16)=136$ USD. For three variables write the three equations in the form $ax+by+cz=d$ with the same variable order in each, for example $x+y+z=10$, $2x+3y+z=17$, $3x+y+4z=31$, giving $x=3$, $y=2$, $z=5$.

A polynomial equation has the form $f(x)=0$ where $f$ is a polynomial, such as $x^{2}+5x-130=0$ or $-4.9t^{2}+15t+12=0$. The solutions are the roots of the equation, which are also the zeros of the function $f$ and the $x$-intercepts of its graph. Use the GDC's polynomial solver, or graph $y=f(x)$ and find where it crosses the $x$-axis. A quadratic can have two, one or no real roots. Write the roots to 3 significant figures: $x^{2}+5x-130=0$ has roots $9.17$ and $-14.2$. To solve $f(x)=k$ for some number $k$, either rearrange to $f(x)-k=0$ or find the intersections of $y=f(x)$ and $y=k$. For the stone, solving $-4.9t^{2}+15t+12=20$ gives $t=0.688$ and $t=2.37$.

A solution of the equation is not always a solution of the problem. Check each root against the context.

• A width, length or time cannot be negative, so reject negative roots. For $x^{2}+5x-130=0$ the root $-14.2$ is rejected, so the width is $9.17$ cm and the length $x+5=14.17$ cm.
• The positive and negative roots of a projectile model: $t=3.72$ s is when the stone lands; $t=-0.658$ s is before it was thrown.
• Give units and answer the question: perimeter $=2(9.17+14.17)=46.7$ cm. Say why you reject a root in a sentence, because the reason earns a mark.

To find the coefficients of a quadratic $h=at^{2}+bt+c$ you need three pieces of data. Each point gives one linear equation in $a$, $b$ and $c$. A stone is at $22.1$ m when $t=1$, $22.4$ m when $t=2$ and $12.9$ m when $t=3$: $a+b+c=22.1,\qquad4a+2b+c=22.4,\qquad9a+3b+c=12.9.$ Solve with technology: $a=-4.9$, $b=15$, $c=12$. Interpret the numbers: $c$ is the height at $t=0$, the starting height of $12$ m. Then use the model: the stone hits the ground when $h=0$, which gives $t=3.72$ s (after rejecting the negative root). This links to quadratic models (SL 2.5).