Linear parameters, solution counts and inequalities
SAT Topic 9 4:02 English narration · English + 中文 subtitles burned in
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Transcript
An identity holds for every permitted input.
恒等式对所有允许的输入均成立。
Start with a x plus b equals c x plus d.
从 ax + b = cx + d 开始。
Collect terms: open parenthesis a minus c close parenthesis x equals d minus b.
合并同类项：(a - c)x = d - b。
Inspect a minus c before dividing.
在相除之前，先检查 a - c。
If a equals c and b equals d, every real input works.
若 a = c 且 b = d，则所有实数输入均满足方程。
If only a equals c, no input works.
若仅 a = c，则无解。
Otherwise, x equals d minus b, all over a minus c, gives one solution.
否则，x = (d - b) / (a - c) 给出唯一解。
Each task gives its own data.
每个任务都有其自身的数据。
Do not copy the worked example’s numbers.
请勿照搬示例中的数值。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
Collect terms to get p minus 3, times x, equals p minus 3.
合并同类项得到 p 减三乘 x 等于 p 减三。
At p equals 3, the equation is 0 equals 0, so every real x works.
当 p 等于三时，方程是零等于零，所以每个实数 x 都成立。
When p is not 3, divide by p minus 3 to get x equals 1.
当 p 不等于三时，两边除以 p 减三，得到 x 等于一。
There is no value of p giving no solution.
不存在使方程无解的 p。
Substituting x equals 1 gives p plus 2 on both sides.
把 x 等于一代入，两边都得到 p 加二。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请结合前几页的完整背景信息进行分析。
Use cost equals the fixed fee plus the rate times time.
使用“费用等于固定费用加费率乘时间”。
The equations are 31 equals f plus 3 r, and 59 equals f plus 7 r.
两个方程是三十一等于 f 加三 r，以及五十九等于 f 加七 r。
Subtracting, the rate is 59 minus 31 yuan, over 7 minus 3 hours, which is 7 yuan per hour.
相减得费率等于五十九减三十一元，除以七减三小时，即每小时七元。
Then the fee is 31 yuan minus 7 times 3, which is 10 yuan.
于是固定费用等于三十一元减七乘三，即十元。
The budget, 10 plus 7 t at most 80, gives t at most 10 hours.
预算条件十加七 t 不超过八十，得到 t 不超过十小时。
Ten whole hours is affordable; eleven costs 87 yuan.
十个整小时负担得起；十一小时要花八十七元。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请结合前几页的完整背景信息。
The first boundary is dashed and the second is solid.
第一条边界是虚线，第二条是实线。
The common region is above the first line, and on or below the second.
公共区域在第一条直线上方，并在第二条直线下方或线上。
At 2 comma 3, the first condition reads 3 greater than 3, which fails.
在点（二，三），第一个条件变成三大于三，不成立。
For the vertical interval to exist, 2 x minus 1 must be less than negative x plus 5, so x is less than 2.
要让竖直方向上的区间存在，必须有二 x 减一小于负 x 加五，所以 x 小于二。
The lines meet at 2 comma 3, but their meeting point is excluded.
两条直线交于（二，三），但交点本身不包括在内。
Return to the task and explain why the method fits.
回到该任务，并解释为何此方法适用。
Original practice does not predict an official score.
原始练习无法预测正式成绩。
Complete required live or recorded evidence separately.
请单独完成所需的现场或录制证据。
Pause and test yourself. Cover the worked answer, rebuild the answer to the last example in your own words, and name one condition that would change it before you check.
暂停一下，自我检测：遮住示范答案，用自己的话重做最后一个例子的答案，并在核对之前说出一个会改变结论的条件。