Exam 10: Systems of Linear Equations

A certain number has three digits and the sum of those digits 12 . When the number is divided by the number formed by the hundreds and the ones digits, the answer equals 11 . In addition, the sum of the hundreds and the tens digits is 10 bigger than the ones digit. What is the number?

Solve by any method: $3.2 x-1.6 y=7.0$ $1.9 x+2.5 y=6.6$

Solve by any algebraic method: $5 y+x=21$ $3 y-x=3$

Solve: $\frac{x-1}{y-2}=\frac{4}{5}$ = =

Solve: $x+z=5 a$ y-z=-6a 2x+3y=-4a

Solve simultaneously: $\frac{10}{x}+\frac{45}{y}=-4$ $\frac{26}{x}+\frac{10}{y}=11$

The sum of the digits of a certain three-digit number is nine. The sum of the hundreds and the tens digits is equal to the ones digit minus one. The number, divided by nine, equals three times the ones digit. What is the number?

Solve: $2 x+4 y+z=5$ $x+y+z=6$ $2 x+3 y+z=6$

Solve simultaneously: $\frac{2 x}{5}+\frac{5 y}{6}=25$ $x-\frac{4 y}{9}=17$

Solve by any method: $2.75 x-3.43 y=-4.144$ $3.30 x+2.50 y=8.59$

Solve by any method: $5 m=6 n-31$ $n=11+2 m$

Solve: $\frac{x}{3}-\frac{y}{2}+\frac{z}{6}=-4$ +-=29 +-=11

The atomic mass of three compounds are known: methane $\left(\mathrm{CH}_{4}\right)$ is 16.0 , glycerol $(\mathrm{C} 3 \mathrm{H} 8 \mathrm{O} 3)$ is 91.9 , and water $(\mathrm{H} 2 \mathrm{O})$ is 18.0. Calculate the atomic mass of $\mathbf{C}, \mathbf{H}$ , and $\mathbf{O}$ by solving this linear system: $\mathbf{C}+4 \mathbf{H}=16.0$ 3+8+3=91.9 2+=18.0

Solve for $x$ and $y$ in terms of the other literal quantities: $2 x+3 y=a$ $5 x-y=b$

Solve by any method: $-2.5 x-3.4 y=-7.4$ $2.1 x+2.5 y=6.9$

Solve simultaneously. Keep 3 significant digits in your final answers, but do not round until the end: += -=

Solve for currents $I_{1}$ and $I_{2}$ in the parallel circuit: $100 I_{1}+35 I_{1}-35 I_{2}=12.0$ $40 I_{2}+35 I_{2}-35 I_{1}=18.0$

Solve by any method: $4.2 x+0.91 y=16$ $3.0 x-4.1 y=11$

Solve. Leave your answers in fractional form: $3 x+4 y-z=1$ 2x-2y+z=0 x+y-z=2

Solve: $x+z=35$ $x+y=25$ $y+z=32$