Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

Notes

Problems & Video

\(\textbf{1)}\) Use Cramer’s Rule to solve the following system of equations.
                    \(3x+4y=18\)
                    \(2x-y=1\)

Show Answer

The answer is \(x=2,y=3\)

\(\textbf{2)}\) \(x-4y=12\)
\(\hspace{11pt}x+3y=-2\)

Show Answer

The answer is \((4,-2)\)

Show Work

\(\,\,\,\,\,x=\frac{\left|{\begin{array}{cc}
12 & -4 \\
-2 & 3 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -4 \\
1 & 3 \\
\end{array} }\right|}=\frac{(12)(3)-(-4)(-2)}{(1)(3)-(-4)(1)}=\frac{36-8}{3+4}=\frac{28}{7}=4\)

\(\,\,\,\,\,y=\frac{\left|{\begin{array}{cc}
1 & 12 \\
1 & -2 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -4 \\
1 & 3 \\
\end{array} }\right|}=\frac{(1)(-2)-(12)(1)}{(1)(3)-(-4)(1)}=\frac{-2-12}{3+4}=\frac{-14}{7}=-2\)

\(\,\,\,\,\,\text{The answer is } (4,-2)\)

\(\textbf{3)}\) \(2x+3y=11\)
\(\hspace{11pt}x+y=3\)

Show Answer

The answer is \((-2,5)\)

Show Work

\(\,\,\,\,\,x=\frac{\left|{\begin{array}{cc}
11 & 3 \\
3 & 1 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
2 & 3 \\
1 & 1 \\
\end{array} }\right|}=\frac{(11)(1)-(3)(3)}{(2)(1)-(3)(1)}=\frac{11-9}{2-3}=\frac{2}{-1}=-2\)

\(\,\,\,\,\,y=\frac{\left|{\begin{array}{cc}
2 & 11 \\
1 & 3 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
2 & 3 \\
1 & 1 \\
\end{array} }\right|}=\frac{(2)(3)-(1)(11)}{(2)(1)-(3)(1)}=\frac{6-11}{2-3}=\frac{-5}{-1}=5\)

\(\,\,\,\,\,\text{The answer is } (-2,5)\)

\(\textbf{4)}\) \(x-y=-6\)
\(\hspace{11pt}5x+2y=12\)

Show Answer

The answer is \((0,6)\)

Show Work

\(\,\,\,\,\,x=\frac{\left|{\begin{array}{cc}
-6 & -1 \\
12 & 2 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -1 \\
5 & 2 \\
\end{array} }\right|}=\frac{(-6)(2)-(-1)(12)}{(1)(2)-(-1)(5)}=\frac{-12+12}{2+5}=\frac{0}{7}=0\)

\(\,\,\,\,\,y=\frac{\left|{\begin{array}{cc}
1 & -6 \\
5 & 12 \\
\end{array} }\right|}{\left|{\begin{array}{cc}
1 & -1 \\
5 & 2 \\
\end{array} }\right|}=\frac{(1)(12)-(-6)(5)}{(1)(2)-(-1)(5)}=\frac{12+30}{2+5}=\frac{42}{7}=6\)

\(\,\,\,\,\,\text{The answer is } (0,6)\)

\(\textbf{5)}\) \(7x+4y=31\)
\(\hspace{11pt}3x+2y=15\)

Show Answer

The answer is \((1,6)\)

\(\textbf{6)}\) \(5x-4y=9\)
\(\hspace{11pt}7x+3y=4\)

Show Answer

The answer is \((1,-1)\)

\(\textbf{7)}\) \(15x-3y=126\)
\(\hspace{11pt}2x+5y=33\)

Show Answer

The answer is \((9,3)\)

\(\textbf{8)}\) \(2x-y=-1\)
\(\hspace{11pt}3x+4y=-7\)

Show Answer

The answer is \((-1,-1)\)

\(\textbf{9)}\) \(2x-7y=-18\)
\(\hspace{11pt}3x+8y=47\)

Show Answer

The answer is \((5,4)\)

\(\textbf{10)}\) \(3x+2y=12\)
\(\hspace{15pt}5x-4y=-2\)

Show Answer

The answer is \((2,3)\)

\(\textbf{11)}\) \(3x+y=10\)
\(\hspace{11pt}x-y=2\)

Show Answer

The answer is \((3,1)\)

See Related Pages\(\)

\(\bullet\text{ Intro to Matrices}\)
\(\,\,\,\,\,\,\,\,\)\(\left[ {\begin{array}{ccc}4 & -5 & 2 \\1 & 0 & 3 \\\end{array} } \right]\)

\(\bullet\text{ Matrix Operations}\)
\(\,\,\,\,\,\,\,\,\)\( \left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)\(+\left[ {\begin{array}{ccc}3 & 45 & 6 \\-8 & 2 & 4 \\1 & 0 & 3 \\\end{array} } \right]\)

\(\bullet\text{ Multiplying Matrices}\)
\(\,\,\,\,\,\,\,\,\)\(\left[{\begin{array}{cc} 1 & 2 \\ -3 & -4 \\\end{array} } \right]\)\(\left[ {\begin{array}{cc}6 & -3 \\5 & 0 \\\end{array} } \right]\)

\(\bullet\text{ Determinants}\)
\(\,\,\,\,\,\,\,\,\)\(\left|{\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right|=ad-bc\)

\(\bullet\text{ Cramer’s Rule}\)
\(\,\,\,\,\,\,\,\,\text{ax+by=e } \& \text{ cx+dy=f}…\)

\(\bullet\text{ Identity Matrix}\)
\(\,\,\,\,\,\,\,\,\)\(\left[{\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]\)

\(\bullet\text{ Identity and Inverse Matrices}\)
\(\,\,\,\,\,\,\,\,A^{-1}=\displaystyle\frac{1}{ad-bc}\left[{\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right]\)

\(\bullet\text{ Transpose Matrix}\)
\(\,\,\,\,\,\,\,\,\left[{\begin{array}{ccc} 1 \\ 2 \\ 5 \\ \end{array} } \right]\Rightarrow\left[{\begin{array}{c} 1 & 2 & 5 \end{array} } \right]\)

\(\bullet\text{ Rotation Matrix}\)
\(\,\,\,\,\,\,\,\,\)\(R(\theta)=\left[{\begin{array}{cc}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \\\end{array} } \right]\)

\(\bullet\text{ Eigenvectors and Eigenvalues}\)
\(\,\,\,\,\,\,\,\,(A-\lambda I)\vec{v}=\vec{0}\)

About Andymath.com

Andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. If you have any requests for additional content, please contact Andy at tutoring@andymath.com. He will promptly add the content.

Topics cover Elementary Math, Middle School, Algebra, Geometry, Algebra 2/Pre-calculus/Trig, Calculus and Probability/Statistics. In the future, I hope to add Physics and Linear Algebra content.

Visit me on Youtube, Tiktok, Instagram and Facebook. Andymath content has a unique approach to presenting mathematics. The clear explanations, strong visuals mixed with dry humor regularly get millions of views. We are open to collaborations of all types, please contact Andy at tutoring@andymath.com for all enquiries. To offer financial support, visit my Patreon page. Let’s help students understand the math way of thinking!

Thank you for visiting. How exciting!