Linear Systems Review

Before the test, we did a quick warmup at the whiteboards in random groups. Each group worked through three task cards, one question of each kind from the unit: a system to solve by substitution, a system to solve by elimination, and a word problem.

There were five versions of the cards, so each group had different numbers. The lettered parts of each question below are those five versions.

If you were away

Choose one part of each question, and work through the three questions in order before looking at any answers. If you’d like more practice, try another part.

Once you have an answer, expand the answer box to check your work. If a question didn’t go the way you expected, try to explain why before moving on.

TIP

A quick way to be sure of your answer is to substitute it into both equations and confirm that the left side equals the right side.

Question 1

Solve the system. Use the method that fits best.

a)

$\begin{array}{rlrl}& \text{A}:& y& =2x-7\\ & \text{B}:& 3x+2y& =7\end{array}$

b)

$\begin{array}{rlrl}& \text{A}:& x& =3y+5\\ & \text{B}:& 2x+5y& =-1\end{array}$

c)

$\begin{array}{rlrl}& \text{A}:& y& =-3x+4\\ & \text{B}:& 5x+2y& =6\end{array}$

d)

$\begin{array}{rlrl}& \text{A}:& x& =4-2y\\ & \text{B}:& 3x-4y& =22\end{array}$

e)

$\begin{array}{rlrl}& \text{A}:& y& =4x+9\\ & \text{B}:& 2x-3y& =3\end{array}$

Answers

In every part, one variable is already isolated in equation A, so substitution fits best.

a) $(3,-1)$. Substitute $y=2x-7$ into B: $3x+2(2x-7)=7$, so $7x-14=7$ and $x=3$. Then $y=2(3)-7=-1$. Check in B: $3(3)+2(-1)=7$.

b) $(2,-1)$. Substitute $x=3y+5$ into B: $2(3y+5)+5y=-1$, so $11y+10=-1$ and $y=-1$. Then $x=3(-1)+5=2$. Check in B: $2(2)+5(-1)=-1$.

c) $(2,-2)$. Substitute $y=-3x+4$ into B: $5x+2(-3x+4)=6$, so $-x+8=6$ and $x=2$. Then $y=-3(2)+4=-2$. Check in B: $5(2)+2(-2)=6$.

d) $(6,-1)$. Substitute $x=4-2y$ into B: $3(4-2y)-4y=22$, so $12-10y=22$ and $y=-1$. Then $x=4-2(-1)=6$. Check in B: $3(6)-4(-1)=22$.

e) $(-3,-3)$. Substitute $y=4x+9$ into B: $2x-3(4x+9)=3$, so $-10x-27=3$ and $x=-3$. Then $y=4(-3)+9=-3$. Check in B: $2(-3)-3(-3)=3$.

Question 2

Solve the system. Use the method that fits best.

a)

$\begin{array}{rlrl}& \text{A}:& 3x+4y& =10\\ & \text{B}:& 5x-3y& =7\end{array}$

b)

$\begin{array}{rlrl}& \text{A}:& 2x+5y& =4\\ & \text{B}:& 3x+2y& =-5\end{array}$

c)

$\begin{array}{rlrl}& \text{A}:& 4x-3y& =18\\ & \text{B}:& 3x+5y& =-1\end{array}$

d)

$\begin{array}{rlrl}& \text{A}:& 5x+2y& =-4\\ & \text{B}:& 3x-7y& =-27\end{array}$

e)

$\begin{array}{rlrl}& \text{A}:& 6x+5y& =14\\ & \text{B}:& 4x-3y& =-16\end{array}$

Answers

Both equations are in standard form and no coefficient is $1$, so elimination fits best. You need to multiply both equations to match a pair of coefficients.

a) $(2,1)$. Multiply A by $3$ and B by $4$: $9x+12y=30$ and $20x-12y=28$. Add: $29x=58$, so $x=2$. Then $3(2)+4y=10$ gives $y=1$. Check in B: $5(2)-3(1)=7$.

b) $(-3,2)$. Multiply A by $3$ and B by $2$: $6x+15y=12$ and $6x+4y=-10$. Subtract: $11y=22$, so $y=2$. Then $2x+5(2)=4$ gives $x=-3$. Check in B: $3(-3)+2(2)=-5$.

c) $(3,-2)$. Multiply A by $5$ and B by $3$: $20x-15y=90$ and $9x+15y=-3$. Add: $29x=87$, so $x=3$. Then $4(3)-3y=18$ gives $y=-2$. Check in B: $3(3)+5(-2)=-1$.

d) $(-2,3)$. Multiply A by $7$ and B by $2$: $35x+14y=-28$ and $6x-14y=-54$. Add: $41x=-82$, so $x=-2$. Then $5(-2)+2y=-4$ gives $y=3$. Check in B: $3(-2)-7(3)=-27$.

e) $(-1,4)$. Multiply A by $2$ and B by $3$: $12x+10y=28$ and $12x-9y=-48$. Subtract: $19y=76$, so $y=4$. Then $6x+5(4)=14$ gives $x=-1$. Check in B: $4(-1)-3(4)=-16$.

Question 3

Write “let” statements and two equations, then solve.

a) Tickets to the school play cost $12 for adults and $7 for students. In total, 280 tickets were sold, and the total revenue was $2560. How many tickets of each kind were sold?

b) Maya invested $8000 in total. She invested part of it at 4% and the rest at 6%. After one year, she had earned $410 in interest altogether. How much did Maya invest at each rate?

c) Flying with the wind, a plane travels 2400 km in 4 hours. The return trip, against the same wind, takes 5 hours. Find the speed of the plane in still air and the speed of the wind.

d) A taxi company charges a fixed fee plus a rate for each kilometre travelled. A 10 km ride costs $23.50. A 16 km ride costs $34.30. Find the fixed fee and the rate per kilometre.

e) A jar holds 85 coins. Every coin is either a loonie ($1) or a toonie ($2). Altogether, the coins are worth $127. How many coins of each kind are in the jar?

Answers

a) 120 adult tickets and 160 student tickets.

Let $a$ be the number of adult tickets and $s$ the number of student tickets. One equation counts tickets, the other counts money:

$\begin{array}{rl}a+s& =280\\ 12a+7s& =2560\end{array}$

Multiply the first equation by $7$ and subtract it from the second: $5a=600$, so $a=120$ and $s=160$. Check: $12(120)+7(160)=1440+1120=2560$.

b) $3500 at 4% and $4500 at 6%.

Let $x$ be the amount, in dollars, invested at 4% and $y$ the amount invested at 6%. One equation is for the money invested, the other for the interest:

$\begin{array}{rl}x+y& =8000\\ 0.04x+0.06y& =410\end{array}$

Multiply the second equation by $100$ to get $4x+6y=41000$, and the first by $4$ to get $4x+4y=32000$. Subtract: $2y=9000$, so $y=4500$ and $x=3500$. Check: $0.04(3500)+0.06(4500)=140+270=410$.

c) The plane flies at 540 km/h in still air, and the wind blows at 60 km/h.

Let $p$ be the speed of the plane in still air and $w$ the speed of the wind, both in km/h. Speed is distance divided by time, so the plane goes $2400÷4=600$ km/h with the wind and $2400÷5=480$ km/h against it:

$\begin{array}{rl}p+w& =600\\ p-w& =480\end{array}$

Add: $2p=1080$, so $p=540$ and $w=60$. Check: $4(540+60)=2400$ and $5(540-60)=2400$.

d) The fixed fee is $5.50 and the rate is $1.80 per kilometre.

Let $f$ be the fixed fee and $r$ the rate per kilometre, both in dollars. The fee is charged once, so it is not multiplied by the distance:

$\begin{array}{rl}f+10r& =23.50\\ f+16r& =34.30\end{array}$

Subtract the first equation from the second: $6r=10.80$, so $r=1.80$. Then $f=23.50-10(1.80)=5.50$. Check: $5.50+16(1.80)=34.30$.

e) 43 loonies and 42 toonies.

Let $l$ be the number of loonies and $t$ the number of toonies. One equation counts coins, the other counts value:

$\begin{array}{rl}l+t& =85\\ l+2t& =127\end{array}$

Subtract the first equation from the second: $t=42$, so $l=43$. Check: $43+2(42)=127$.

Curriculum connection

B1.1

solve systems of two linear equations involving two variables, using the algebraic method of substitution or elimination (Sample problem: Solve $y=\frac{1}{2}x-5$, $3x+2y=-2$ for $x$ and $y$ algebraically, and verify algebraically and graphically);

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B1.2

solve problems that arise from realistic situations described in words or represented by linear systems of two equations involving two variables, by choosing an appropriate algebraic or graphical method (Sample problem: The Robotics Club raised $5000 to build a robot for a future competition. The club invested part of the money in an account that paid 4% annual interest, and the rest in a government bond that paid 3.5% simple interest per year. After one year, the club earned a total of $190 in interest. How much was invested at each rate? Verify your result.).

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