Exam 10: Systems of Linear Equations

Solve simultaneously: $\frac{x}{3}-\frac{y}{4}=2$ $\frac{x}{2}-\frac{y}{3}=4$

Solve. Leave your answers in fractional form: $\frac{1}{x}+\frac{2}{z}=7$ += -+=20

The sum of two numbers is 67 . The quotient of the greater divided by the smaller number is 7 with remainder 3. Determine the two numbers.

Solve by any method: $7.41 x-3.01 y=2.11$ $6.92 x+9.00 y=1.00$

Solve: $\frac{2 x+5}{3}-\frac{y-1}{2}+z=11$ +y+=11 x--=-6

A $10.0 \mathrm{~kg}$ mass and a $19.5 \mathrm{~kg}$ mass are hung from either end of a bar of negligible weight. If the 10.0 $\mathrm{kg}$ is increased to $12.0 \mathrm{~kg}$ , the balance point shifts by $16.2 \mathrm{~cm}$ . How long is the bar?

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

Solve by any method. Leave your answers in fractional form: $4 x+5 y=2$ $2 x-5 y=0$

Solve. Leave your answers in fractional form: $\frac{1}{x}+\frac{2}{y}-\frac{2}{z}=6$ +=7 -=3

Solve simultaneously. Leave your answers in fractional form: $\frac{x}{3}+\frac{y}{4}=6$ $\frac{x}{5}+\frac{y}{2}=1$

Solve simultaneously: $\frac{x}{5}-5 y=15$ $\frac{x}{3}-4 y=12$

Solve simultaneously. Leave your answers in fractional form: $\frac{3 x}{2}+y=4$ $\frac{4 x}{5}+\frac{2 y}{3}=1$

Solve simultaneously. Leave your answers in fractional form: $\frac{2}{x}+\frac{y}{3}=5$ $\frac{1}{x}-y=3$

A projectile is launched from height $h_{o}$ above the ground. Its height (metres) after $t$ seconds is given by $s=k t^{2}+v_{0} t+h_{0}$ where $k$ is a constant, $v_{0}$ is the initial velocity. If the height at $1.0 \mathrm{~s}$ is $7.75 \mathrm{~m}$ and the height at $1.5 \mathrm{~s}$ is $6.625 \mathrm{~m}$ , and the height at $2.0 \mathrm{~s}$ is $5.00 \mathrm{~m}$ , determine the height at time $2.5 \mathrm{~s}$ .

There is a fraction such that if 5 is subtracted from the numerator, its value will be $1 / 4$ , and if 7 is added to the denominator, its value will be $1 / 3$ . What is the fraction?

Solve by any method: $3.52 x-2.47 y=-0.037$ $1.09 x+1.50 y=6.68$

Solve by any algebraic method: $25 x+15 y=5$ $10 x+10 y=10$

If the width of a rectangle is increased by 4 and the length increased by 1 , the area increases by 62 . If instead the width is increased by 2 and the length decreased by 3 , the area decreases by 12 . Find the original dimensions.

Solve by any method: 3x + 2y = 1, x − 3y = −18

Solve. Leave your answers in fractional form: $\frac{x}{3}+\frac{y}{2}+\frac{z}{4}=1$ 2x+3y+2z=4 x++=2