What a solution is
A solution of \[ \[\begin{cases} a_1x+b_1y=c_1\\ a_2x+b_2y=c_2 \end{cases}\] \] is an ordered pair $(x,y)$ satisfying both equations at once. Graphically it is a point where the two lines meet.
Two distinct lines in a plane either cross once or never cross. So a system of two linear equations has exactly one solution, no solution, or infinitely many — there is no “two solutions” case.
Điểm mấu chốt của cả unit: nghiệm là một cặp. Tìm được $x$ mà chưa thay ngược lại để lấy $y$ là mới làm nửa bài — và một nửa số câu của chủ đề này hỏi thứ chỉ có được sau khi biết cả hai số, ví dụ $x+y$ hay $xy$.
Method 1 — substitution
Use it when one variable already stands alone, or when a coefficient is $1$ so isolating costs nothing.
Solve $\;y=2x-3\;$ and $\;4x+3y=21$.
Substitute the first into the second: $4x+3(2x-3)=21$, so $4x+6x-9=21$ and $10x=30$, giving $x=3$. Then $y=2(3)-3=3$.
Solution: $\boxed{(3,3)}$. Check in the second equation: $4(3)+3(3)=21$.
Thế xong ra $x$ rồi khoanh luôn. Đề hỏi $y$, hoặc hỏi $x+y$, hoặc hỏi toạ độ giao điểm — đọc lại câu hỏi sau khi có đáp số, mỗi lần, không có ngoại lệ.
Method 2 — elimination
Scale one or both equations so that one variable cancels when you add or subtract.
Solve $\;3x+4y=10\;$ and $\;5x-2y=8$.
Multiply the second equation by $2$: $10x-4y=16$. Adding to the first equation eliminates $y$: \[ (3x+10x)+(4y-4y)=10+16 \;\Longrightarrow\; 13x=26 \;\Longrightarrow\; x=2 . \] Then $3(2)+4y=10$ gives $4y=4$ and $y=1$.
Solution: $\boxed{(2,1)}$. Check: $5(2)-2(1)=8$.
Solve $\;4x+5y=7\;$ and $\;6x-3y=21$.
The second equation divides by $3$ first: $2x-y=7$, so $y=2x-7$.
Substituting: $4x+5(2x-7)=7$, so $14x=42$ and $x=3$, hence $y=2(3)-7=-1$.
Solution: $\boxed{(3,-1)}$. Check: $4(3)+5(-1)=7$ and $6(3)-3(-1)=21$.
Chọn cách theo hình dạng của hệ, không theo thói quen:
- Một ẩn đã đứng riêng, hoặc có hệ số $1$ $\;\rightarrow\;$ thế.
- Hai phương trình cùng dạng $ax+by=c$ $\;\rightarrow\;$ khử.
- Trước khi làm gì cả, rút gọn: $6x-3y=21$ chia $3$ thành $2x-y=7$, số nhỏ đi ba lần.
Solve $\;\dfrac{x}{2}+\dfrac{y}{3}=4\;$ and $\;\dfrac{x}{4}-\dfrac{y}{6}=1$.
Clear denominators first: multiply the first by $6$ to get $3x+2y=24$, and the second by $12$ to get $3x-2y=12$.
Adding: $6x=36$, so $x=6$. Then $18+2y=24$ gives $y=3$.
Solution: $\boxed{(6,3)}$. Check: $\tfrac62+\tfrac33=3+1=4$ and $\tfrac64-\tfrac36=1.5-0.5=1$.
Solve $\;0.4x+0.3y=2.7\;$ and $\;0.5x-0.2y=0.5$.
Multiply both by $10$: $4x+3y=27$ and $5x-2y=5$.
Multiply the first by $2$ and the second by $3$: $8x+6y=54$ and $15x-6y=15$. Adding gives $23x=69$, so $x=3$, and then $12+3y=27$ gives $y=5$.
Solution: $\boxed{(3,5)}$.
Method 3 — graphing, and what it actually costs
Typing both equations into a graphing calculator and clicking the intersection is a legitimate method, and on the digital test it is often the fastest one. It is also the method that fails most spectacularly when misapplied.
- Coefficients are decimals or awkward numbers, and the answer is a plain number.
- The question asks how many solutions the system has.
- The system already comes as two “$y=\ldots$” lines.
- You have solved algebraically and want a five-second confirmation.
- The system contains a letter such as $a$, $c$ or $k$ — there is nothing to plot.
- The answer is a combination like $x+y$ or $5x-2y$; adding the two equations gets there before the graph finishes rendering.
- The answer is an exact fraction and the intersection reads as a rounded decimal.
- The question asks which statement must be true.
The system $\;y=1.4x-3.2\;$ and $\;y=-0.6x+5.8\;$ has one solution. What is its $x$-coordinate?
This is the shape graphing was built for: two lines already solved for $y$, decimal coefficients, and a numerical answer. Typing both and reading the intersection takes seconds.
By hand it is still short: $1.4x-3.2=-0.6x+5.8$ gives $2x=9$, so $x=\boxed{4.5}$.
Which method is fastest for $\;7x+2y=24\;$ and $\;3x-2y=6$, and what is $y$?
Neither variable is isolated, so substitution would create fractions; the graph would give a decimal to read off. But the $y$-coefficients are already opposites, so one addition finishes it: \[ 10x=30 \;\Longrightarrow\; x=3, \qquad 7(3)+2y=24 \;\Longrightarrow\; 2y=3 \;\Longrightarrow\; y=\boxed{\tfrac32}. \] Check: $3(3)-2\!\left(\tfrac32\right)=9-3=6$.
Quy tắc thực dụng: bấm đồ thị khi đáp án là một con số, giải tay khi đáp án là một biểu thức hoặc có chữ cái. Học sinh mất điểm nhiều nhất không phải vì không biết bấm, mà vì bấm cho những câu mà không có gì để bấm.
Đọc giao điểm trên đồ thị ra $x=1{,}833\ldots$ rồi ghi $1.83$ vào ô tự điền. Đáp số thật là $\tfrac{11}{6}$. Ô tự điền chấp nhận số thập phân nhưng phải đủ độ chính xác — gặp số lặp thì nộp phân số, đừng nộp số làm tròn.
The questions that never need a full solve
This is the highest-yield idea in the unit. Add the two equations. Subtract them. Look at what comes out before deciding to solve anything.
If the question asks for $x+y$, $x-y$, $2x+2y$, $x^2-y^2$ or similar, try $(\text{eq}_1)+(\text{eq}_2)$ and $(\text{eq}_1)-(\text{eq}_2)$ first. Symmetric systems — where the coefficients of the second equation are the first pair swapped — almost always collapse in one step.
If $\;5x+2y=17\;$ and $\;2x+5y=11$, what is the value of $x+y$?
Add the two equations: $7x+7y=28$, so $x+y=\boxed{4}$ — with no need to find $x$ or $y$.
(Subtracting gives $3x-3y=6$, so $x-y=2$; together these give $x=3$, $y=1$, which checks: $5(3)+2(1)=17$ and $2(3)+5(1)=11$.)
If $\;4x+3y=23\;$ and $\;2x+5y=29$, what is the value of $5x-2y$?
Here the target is not symmetric, so solve properly. Double the second equation: $4x+10y=58$. Subtracting the first: $7y=35$, so $y=5$, and then $4x+15=23$ gives $x=2$.
Therefore $5x-2y=10-10=\boxed{0}$.
Note what happens if you stop at $x=2$: you hand in $2$, and lose the point.
Trừ hai phương trình mà chỉ đổi dấu vế trái, quên vế phải. $(4x+10y)-(4x+3y)=58-23$ — vế phải cũng phải trừ. Đây là lỗi số một khi khử bằng phép trừ; nếu hay mắc, hãy nhân một phương trình với $-1$ rồi cộng, không bao giờ trừ.
A letter in the coefficients
If a constant $c$ or $k$ sits in the system, the question is about how many solutions exist, not about their values.
The system $a_1x+b_1y=c_1$, $a_2x+b_2y=c_2$ has exactly one solution precisely when \[ a_1b_2-a_2b_1\neq 0 , \] that is, when the two lines have different slopes. If $a_1b_2-a_2b_1=0$ the system has either no solution or infinitely many, depending on the constants.
For what value of the constant $c$ does the system $\;2x+3y=12\;$ and $\;4x+cy=9\;$ fail to have exactly one solution?
The condition for a unique solution is $2c-4\cdot3\neq0$, that is $2c\neq12$ and $c\neq6$.
So the system fails to have exactly one solution when $\boxed{c=6}$.
(With $c=6$ the second equation is $4x+6y=9$, i.e. $2x+3y=4.5$, which contradicts $2x+3y=12$ — no solution.)
The solution of the system $\;3x+ky=12\;$ and $\;2x-3y=1\;$ has $x=3$. What is the value of $k$?
Put $x=3$ into the second equation: $6-3y=1$, so $3y=5$ and $y=\tfrac53$.
Now use the first: $9+k\cdot\tfrac53=12$, so $\tfrac53k=3$ and $k=\boxed{\tfrac95}$.
Check: $3(3)+\tfrac95\cdot\tfrac53=9+3=12$.
Đề cho sẵn một phần của nghiệm (“the solution has $x=3$”, “the graphs intersect at $x=3$”) là đang tặng bạn một phương trình một ẩn. Thay ngay vào phương trình không chứa chữ cái — phương trình đó luôn giải được sạch — rồi mới quay lại phương trình có tham số.